Integration by Parts Practice Problems with Solutions

15 integration by parts practice problems with step-by-step solutions, from $\int xe^{3x}\,dx$ to logarithms, inverse trigonometric functions, cyclic integrals and definite integrals. Each problem has a hint, then the full solution. New to the method? Start with integration by parts explained: formula, proof and examples.

Contents8 sections

The formula you need

\int u\,dv = uv - \int v\,du,\qquad \int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du.

Choose $u$ with the LIATE order (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential); the rest of the integrand is $dv$ . When the rule must be applied several times, the tabular (DI) method keeps the signs straight.

Basic Integration by Parts: Polynomial–Exponential Products

These exercises introduce the formula $\int u\,dv = uv - \int v\,du$ in its simplest setting: a linear or quadratic polynomial multiplied by an exponential function. Focus on a clean LIATE choice and careful bookkeeping of signs.

Exercise 1: Linear Polynomial Times Exponential

Easy

Evaluate the following indefinite integral.

Find $\displaystyle \int x e^{3x}\,dx$ .
Verify your answer by differentiating the result.

Hint

Let

u = x

(algebraic, simplifies on differentiation) and

dv = e^{3x}\,dx

. Compute

du

and

v

, then apply

\int u\,dv = uv - \int v\,du

. Don’t forget the constant of integration.

View Solution

Solution to question 1:

Set $u = x$ and $dv = e^{3x}\,dx$ . Then $du = dx$ and $v = \dfrac{e^{3x}}{3}$ .

\int x e^{3x}\,dx = \frac{x e^{3x}}{3} - \int \frac{e^{3x}}{3}\,dx = \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} + C.

Solution to question 2:

Differentiate $F(x) = \dfrac{x e^{3x}}{3} - \dfrac{e^{3x}}{9}$ :

F'(x) = \frac{e^{3x} + 3x e^{3x}}{3} - \frac{3e^{3x}}{9} = \frac{e^{3x}}{3} + x e^{3x} - \frac{e^{3x}}{3} = x e^{3x}. \checkmark

Exercise 2: Linear Polynomial Times Cosine

Easy

Evaluate each integral using integration by parts.

Compute $\displaystyle \int x \cos(2x)\,dx$ .
Compute $\displaystyle \int (3x - 1)\sin(x)\,dx$ .

Hint

For both integrals, assign the polynomial factor to

u

(it vanishes after one differentiation) and the trigonometric factor to

dv

. Integrate

dv

carefully, watching the chain-rule factor for

\cos(2x)

View Solution

Solution to question 1:

Let $u = x$ , $dv = \cos(2x)\,dx$ . Then $du = dx$ , $v = \dfrac{\sin(2x)}{2}$ .

\int x\cos(2x)\,dx = \frac{x\sin(2x)}{2} - \int \frac{\sin(2x)}{2}\,dx = \frac{x\sin(2x)}{2} + \frac{\cos(2x)}{4} + C.

Solution to question 2:

Let $u = 3x-1$ , $dv = \sin(x)\,dx$ . Then $du = 3\,dx$ , $v = -\cos(x)$ .

\int (3x-1)\sin(x)\,dx = -(3x-1)\cos(x) + 3\int \cos(x)\,dx = -(3x-1)\cos(x) + 3\sin(x) + C.

Exercise 3: Quadratic Polynomial Times Exponential

Medium

Some polynomial–exponential integrals require integration by parts more than once.

Compute $\displaystyle \int x^2 e^{-x}\,dx$ .
Compute $\displaystyle \int (x^2 + 2x)\,e^{2x}\,dx$ .

Hint

Apply integration by parts with

u = x^2

(or

u = x^2 + 2x

) and

dv = e^{\pm x}\,dx

. After the first application the polynomial degree drops by one; apply the formula a second time to finish.

View Solution

Solution to question 1:

First pass: $u = x^2$ , $dv = e^{-x}\,dx \Rightarrow du = 2x\,dx$ , $v = -e^{-x}$ .

\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int x e^{-x}\,dx.

Second pass: $u = x$ , $dv = e^{-x}\,dx \Rightarrow du = dx$ , $v = -e^{-x}$ .

\int x e^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} - e^{-x} + C.

Combining:

\int x^2 e^{-x}\,dx = -x^2 e^{-x} - 2xe^{-x} - 2e^{-x} + C = -e^{-x}(x^2 + 2x + 2) + C.

Solution to question 2:

Let $u = x^2 + 2x$ , $dv = e^{2x}\,dx \Rightarrow du = (2x+2)\,dx$ , $v = \dfrac{e^{2x}}{2}$ .

\int (x^2+2x)e^{2x}\,dx = \frac{(x^2+2x)e^{2x}}{2} - \int \frac{(2x+2)e^{2x}}{2}\,dx.

Now apply once more with $u = x+1$ , $dv = e^{2x}\,dx$ :

\int (x+1)e^{2x}\,dx = \frac{(x+1)e^{2x}}{2} - \frac{e^{2x}}{4} + C.

Combining:

\int (x^2+2x)e^{2x}\,dx = \frac{(x^2+2x)e^{2x}}{2} - \frac{(x+1)e^{2x}}{2} + \frac{e^{2x}}{4} + C = e^{2x}\!\left(\frac{x^2+x}{2} - \frac{1}{4}\right) + C.

Integration by Parts with Logarithmic Functions

Logarithmic functions occupy the top of the LIATE hierarchy, so they are almost always assigned the role of $u$ . A characteristic trick is writing $\int \ln x\,dx = \int 1 \cdot \ln x\,dx$ to manufacture a $dv$ .

Exercise 4: The Classic Logarithm Integral

Easy

Logarithms appear alone in many integrals; a silent factor of 1 supplies the $dv$ .

Compute $\displaystyle \int \ln x\,dx$ .
Compute $\displaystyle \int \ln(5x)\,dx$ .

Hint

Rewrite as

\int 1 \cdot \ln(\cdot)\,dx

. Let

u = \ln(\cdot)

and

dv = dx

. For part 2, use the logarithm property

\ln(5x) = \ln 5 + \ln x

to simplify, or apply the formula directly.

View Solution

Solution to question 1:

Let $u = \ln x$ , $dv = dx \Rightarrow du = \tfrac{1}{x}\,dx$ , $v = x$ .

\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

Solution to question 2:

Using $\ln(5x) = \ln 5 + \ln x$ :

\int \ln(5x)\,dx = \int \ln 5\,dx + \int \ln x\,dx = x\ln 5 + x\ln x - x + C = x\ln(5x) - x + C.

Exercise 5: Polynomial Factor Paired with a Logarithm

Medium

When a polynomial multiplies a logarithm, integrate the polynomial (assign it to $dv$ ) and differentiate the logarithm.

Compute $\displaystyle \int x^3 \ln x\,dx$ .
Compute $\displaystyle \int x\,\ln(x^2 + 1)\,dx$ .

Hint

For question 1, let

u = \ln x

and

dv = x^3\,dx

. For question 2, first try the substitution

t = x^2 + 1

to see whether integration by parts can be avoided, or directly apply the formula with

u = \ln(x^2+1)

and

dv = x\,dx

View Solution

Solution to question 1:

Let $u = \ln x$ , $dv = x^3\,dx \Rightarrow du = \tfrac{1}{x}\,dx$ , $v = \tfrac{x^4}{4}$ .

\int x^3 \ln x\,dx = \frac{x^4 \ln x}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x}\,dx = \frac{x^4 \ln x}{4} - \frac{1}{4}\int x^3\,dx = \frac{x^4 \ln x}{4} - \frac{x^4}{16} + C.

Solution to question 2:

Let $u = \ln(x^2+1)$ , $dv = x\,dx \Rightarrow du = \tfrac{2x}{x^2+1}\,dx$ , $v = \tfrac{x^2}{2}$ .

\int x\,\ln(x^2+1)\,dx = \frac{x^2}{2}\ln(x^2+1) - \int \frac{x^3}{x^2+1}\,dx.

Perform polynomial division: $\dfrac{x^3}{x^2+1} = x - \dfrac{x}{x^2+1}$ .

\int \frac{x^3}{x^2+1}\,dx = \frac{x^2}{2} - \frac{1}{2}\ln(x^2+1) + C.

Therefore:

\int x\,\ln(x^2+1)\,dx = \frac{x^2}{2}\ln(x^2+1) - \frac{x^2}{2} + \frac{\ln(x^2+1)}{2} + C.

Integration by Parts with Inverse Trigonometric Functions

Inverse trigonometric functions ( $\arctan x$ , $\arcsin x$ , $\arccos x$ ) rank above algebraic functions in the LIATE rule and are not easily integrated directly, making them natural candidates for $u$ . Their derivatives are rational or algebraic, which typically simplifies the residual integral.

Exercise 6: Arctangent and Arcsine Integrals

Easy

Evaluate the following integrals by assigning the inverse trigonometric function to $u$ .

Compute $\displaystyle \int \arctan(x)\,dx$ .
Compute $\displaystyle \int \arcsin(x)\,dx$ .

Hint

In both cases let

u

be the inverse trigonometric function and

dv = dx

. After applying the formula you will obtain a rational or algebraic integral in

v\,du

that is manageable with a standard substitution.

View Solution

Solution to question 1:

Let $u = \arctan x$ , $dv = dx \Rightarrow du = \tfrac{dx}{1+x^2}$ , $v = x$ .

\int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C.

Solution to question 2:

Let $u = \arcsin x$ , $dv = dx \Rightarrow du = \tfrac{dx}{\sqrt{1-x^2}}$ , $v = x$ .

\int \arcsin x\,dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,dx.

Substitute $t = 1 - x^2$ , $dt = -2x\,dx$ :

\int \frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2} + C.

Therefore:

\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C.

Exercise 7: Polynomial Times Inverse Trigonometric Function

Medium

Combining a polynomial with an inverse trig function creates a richer residual integral.

Compute $\displaystyle \int x\,\arctan(x)\,dx$ .
Compute $\displaystyle \int x^2\,\arcsin(x)\,dx$ .

Hint

Let

u

be the inverse trig part and

dv

be the polynomial times

dx

. After differentiating

u

and integrating

dv

, the remaining integral will be a rational function (question 1) or require a trigonometric identity (question 2).

View Solution

Solution to question 1:

Let $u = \arctan x$ , $dv = x\,dx \Rightarrow du = \tfrac{dx}{1+x^2}$ , $v = \tfrac{x^2}{2}$ .

\int x\,\arctan x\,dx = \frac{x^2}{2}\arctan x - \frac{1}{2}\int \frac{x^2}{1+x^2}\,dx.

Since $\dfrac{x^2}{1+x^2} = 1 - \dfrac{1}{1+x^2}$ :

\int x\,\arctan x\,dx = \frac{x^2}{2}\arctan x - \frac{x}{2} + \frac{\arctan x}{2} + C = \frac{x^2+1}{2}\arctan x - \frac{x}{2} + C.

Solution to question 2:

Let $u = \arcsin x$ , $dv = x^2\,dx \Rightarrow du = \tfrac{dx}{\sqrt{1-x^2}}$ , $v = \tfrac{x^3}{3}$ .

\int x^2 \arcsin x\,dx = \frac{x^3}{3}\arcsin x - \frac{1}{3}\int \frac{x^3}{\sqrt{1-x^2}}\,dx.

For the residual integral, write $x^3 = x^2 \cdot x$ and substitute $t = 1 - x^2$ , so $x^2 = 1-t$ and $x\,dx = -\tfrac{dt}{2}$ :

\int \frac{x^3}{\sqrt{1-x^2}}\,dx = \int \frac{(1-t)}{\sqrt{t}}\cdot\left(-\frac{1}{2}\right)dt = \frac{1}{2}\int\!\left(t^{1/2} - t^{-1/2}\right)dt = \frac{(1-x^2)^{3/2}}{3} - \sqrt{1-x^2} + C.

Combining:

\int x^2 \arcsin x\,dx = \frac{x^3}{3}\arcsin x - \frac{(1-x^2)^{3/2}}{9} + \frac{\sqrt{1-x^2}}{3} + C.

Repeated Integration by Parts and the Tabular Method

When a polynomial of degree $n \geq 2$ multiplies a function whose repeated integrals are easily written, applying integration by parts $n$ times in a systematic tabular scheme reduces computation significantly. The tabular (or DI) method organises successive derivatives of $u$ and successive antiderivatives of $dv$ in a table, with alternating signs.

Exercise 8: Applying the Tabular Method

Medium

Use the tabular method to evaluate the following integrals efficiently.

Compute $\displaystyle \int x^3 e^{x}\,dx$ .
Compute $\displaystyle \int x^2 \cos(x)\,dx$ .

Hint

Build a two-column table. In the left column place

x^n

and its successive derivatives (which eventually reach 0). In the right column place the repeatedly integrated trigonometric or exponential function. Multiply diagonally down with alternating signs

+, -, +, \ldots

View Solution

Solution to question 1:

Sign	Derivatives of $x^3$	Antiderivatives of $e^x$
$+$	$x^3$	$e^x$
$-$	$3x^2$	$e^x$
$+$	$6x$	$e^x$
$-$	$6$	$e^x$
$+$	$0$	$e^x$

\int x^3 e^x\,dx = e^x(x^3 - 3x^2 + 6x - 6) + C.

Solution to question 2:

Sign	Derivatives of $x^2$	Antiderivatives of $\cos x$
$+$	$x^2$	$\sin x$
$-$	$2x$	$-\cos x$
$+$	$2$	$-\sin x$
$-$	$0$	$\cos x$

\int x^2 \cos x\,dx = x^2 \sin x + 2x\cos x - 2\sin x + C.

Exercise 9: Higher-Degree Polynomial with Trigonometric Function

Hard

This exercise extends the tabular method to a higher-degree polynomial.

Compute $\displaystyle \int x^4 \sin(x)\,dx$ .
Compute $\displaystyle \int (2x^3 - x)\cos(3x)\,dx$ .

Hint

For question 1 the tabular table will have 5 derivative rows before hitting 0. For question 2, remember that

\dfrac{d^n}{dx^n}(2x^3 - x)

hits 0 after the 4th derivative, and track the chain-rule factors introduced by the

\cos(3x)

column carefully.

View Solution

Solution to question 1:

Sign	Deriv. of $x^4$	Antideriv. of $\sin x$
$+$	$x^4$	$-\cos x$
$-$	$4x^3$	$-\sin x$
$+$	$12x^2$	$\cos x$
$-$	$24x$	$\sin x$
$+$	$24$	$-\cos x$
	$0$

\int x^4 \sin x\,dx = -x^4\cos x + 4x^3\sin x + 12x^2\cos x - 24x\sin x - 24\cos x + C.

Solution to question 2:

Let $p(x) = 2x^3 - x$ . Build the table with $dv = \cos(3x)\,dx$ , so antiderivatives cycle through $\tfrac{\sin(3x)}{3}, -\tfrac{\cos(3x)}{9}, -\tfrac{\sin(3x)}{27}, \tfrac{\cos(3x)}{81}$ .

\int (2x^3-x)\cos(3x)\,dx = \frac{(2x^3-x)\sin(3x)}{3} + \frac{(6x^2-1)\cos(3x)}{9} - \frac{12x\sin(3x)}{27} - \frac{12\cos(3x)}{81} + C.

Simplifying the last term: $\dfrac{12}{81} = \dfrac{4}{27}$ .

= \frac{(2x^3-x)\sin(3x)}{3} + \frac{(6x^2-1)\cos(3x)}{9} - \frac{4x\sin(3x)}{9} - \frac{4\cos(3x)}{27} + C.

Cyclic Integrals: Integration by Parts Applied Twice

When both factors in a product are “permanent” (neither differentiates to zero nor integrates to zero), applying integration by parts twice returns the original integral on the right-hand side. Solving the resulting equation for the integral yields the answer. This technique is central to integrals of the form $\int e^{ax}\sin(bx)\,dx$ or $\int e^{ax}\cos(bx)\,dx$ .

Exercise 10: Exponential Times Sine or Cosine

Medium

Evaluate the following integrals using the cyclic technique.

Compute $\displaystyle \int e^{x}\sin(x)\,dx$ .
Compute $\displaystyle \int e^{2x}\cos(3x)\,dx$ .

Hint

Apply integration by parts twice, keeping the same assignment convention each time (e.g., always let

u = e^{ax}

). After the second application you will see the original integral

I

appear on the right-hand side. Solve the resulting equation for

I

View Solution

Solution to question 1:

Let $I = \int e^x \sin x\,dx$ . Set $u = e^x$ , $dv = \sin x\,dx$ :

I = -e^x\cos x + \int e^x\cos x\,dx.

Apply again with $u = e^x$ , $dv = \cos x\,dx$ :

I = -e^x\cos x + e^x\sin x - \int e^x\sin x\,dx = -e^x\cos x + e^x\sin x - I.

Solving:

2I = e^x(\sin x - \cos x) \implies \boxed{I = \frac{e^x(\sin x - \cos x)}{2} + C}.

Solution to question 2:

Let $I = \int e^{2x}\cos(3x)\,dx$ . Let $u = e^{2x}$ , $dv = \cos(3x)\,dx$ :

I = \frac{e^{2x}\sin(3x)}{3} - \frac{2}{3}\int e^{2x}\sin(3x)\,dx.

Apply again with $u = e^{2x}$ , $dv = \sin(3x)\,dx$ :

\int e^{2x}\sin(3x)\,dx = -\frac{e^{2x}\cos(3x)}{3} + \frac{2}{3}\int e^{2x}\cos(3x)\,dx = -\frac{e^{2x}\cos(3x)}{3} + \frac{2I}{3}.

Substituting back:

I = \frac{e^{2x}\sin(3x)}{3} - \frac{2}{3}\!\left(-\frac{e^{2x}\cos(3x)}{3} + \frac{2I}{3}\right) = \frac{e^{2x}\sin(3x)}{3} + \frac{2e^{2x}\cos(3x)}{9} - \frac{4I}{9}.

\frac{13I}{9} = \frac{3e^{2x}\sin(3x) + 2e^{2x}\cos(3x)}{9} \implies \boxed{I = \frac{e^{2x}(3\sin(3x) + 2\cos(3x))}{13} + C}.

Definite Integrals Using Integration by Parts

When the limits of integration are specified, the boundary term $\bigl[uv\bigr]_a^b$ must be evaluated explicitly. It is essential to apply the limits to both the boundary term and the remaining integral. Always double-check the evaluation of $uv$ at both endpoints.

Exercise 11: Definite Integral with a Polynomial–Exponential Product

Medium

Evaluate the following definite integrals exactly.

Compute $\displaystyle \int_0^1 x e^{x}\,dx$ .
Compute $\displaystyle \int_0^{\pi} x \sin(x)\,dx$ .

Hint

Find the antiderivative using integration by parts, then apply the limits. Remember the formula

\bigl[uv\bigr]_a^b - \int_a^b v\,du

. Evaluate both the boundary term and the definite integral of

v\,du

at the endpoints.

View Solution

Solution to question 1:

Let $u = x$ , $dv = e^x\,dx \Rightarrow du = dx$ , $v = e^x$ .

\int_0^1 x e^x\,dx = \bigl[xe^x\bigr]_0^1 - \int_0^1 e^x\,dx = (1\cdot e - 0) - \bigl[e^x\bigr]_0^1 = e - (e - 1) = 1.

Solution to question 2:

Let $u = x$ , $dv = \sin(x)\,dx \Rightarrow du = dx$ , $v = -\cos(x)$ .

\int_0^{\pi} x\sin(x)\,dx = \bigl[-x\cos(x)\bigr]_0^{\pi} + \int_0^{\pi}\cos(x)\,dx = (-\pi\cos\pi + 0) + \bigl[\sin x\bigr]_0^{\pi}.

= \pi + (\sin\pi - \sin 0) = \pi + 0 = \pi.

Exercise 12: Definite Integral Involving a Logarithm

Medium

Evaluate each definite integral, simplifying your answer fully.

Compute $\displaystyle \int_1^e x\,\ln(x)\,dx$ .
Compute $\displaystyle \int_1^e \ln(x)\,dx$ .

Hint

For question 1, let

u = \ln x

and

dv = x\,dx

. For question 2, use the silent-1 trick. In both cases carefully substitute

x = e

and

x = 1

, recalling that

\ln(e) = 1

and

\ln(1) = 0

View Solution

Solution to question 1:

Let $u = \ln x$ , $dv = x\,dx \Rightarrow du = \tfrac{dx}{x}$ , $v = \tfrac{x^2}{2}$ .

\int_1^e x\ln x\,dx = \left[\frac{x^2\ln x}{2}\right]_1^e - \int_1^e \frac{x}{2}\,dx = \frac{e^2}{2} - 0 - \left[\frac{x^2}{4}\right]_1^e = \frac{e^2}{2} - \frac{e^2-1}{4} = \frac{e^2+1}{4}.

Solution to question 2:

Using the antiderivative from Exercise 4: $\int \ln x\,dx = x\ln x - x$ .

\int_1^e \ln x\,dx = \bigl[x\ln x - x\bigr]_1^e = (e\cdot 1 - e) - (1\cdot 0 - 1) = 0 + 1 = 1.

Advanced and Mixed-Strategy Problems

At this level, a single integral may require a combination of techniques: a preliminary substitution followed by integration by parts, or repeated integration by parts leading to a reduction formula. Recognising which strategy to apply, and in what order,is the mark of a confident practitioner.

Exercise 13: Substitution Followed by Integration by Parts

Hard

These integrals require a substitution before integration by parts can be applied.

Compute $\displaystyle \int \sqrt{x}\,e^{\sqrt{x}}\,dx$ .
Compute $\displaystyle \int \sin(\ln x)\,dx$ .

Hint

For question 1, substitute

t = \sqrt{x}

to eliminate the square root, then apply integration by parts (it may need to be applied twice). For question 2, let

t = \ln x

to convert the integral to

\int e^t \sin(t)\,dt

, which is a cyclic integral.

View Solution

Solution to question 1:

Let $t = \sqrt{x}$ , so $x = t^2$ and $dx = 2t\,dt$ .

\int \sqrt{x}\,e^{\sqrt{x}}\,dx = \int t\,e^t\cdot 2t\,dt = 2\int t^2 e^t\,dt.

Using the tabular method on $2\int t^2 e^t\,dt$ :

2\int t^2 e^t\,dt = 2e^t(t^2 - 2t + 2) + C.

Substituting back $t = \sqrt{x}$ :

\int \sqrt{x}\,e^{\sqrt{x}}\,dx = 2e^{\sqrt{x}}(x - 2\sqrt{x} + 2) + C.

Solution to question 2:

Let $t = \ln x$ , so $x = e^t$ and $dx = e^t\,dt$ .

\int \sin(\ln x)\,dx = \int e^t\sin(t)\,dt.

This is the cyclic integral from Exercise 10, question 1:

\int e^t\sin(t)\,dt = \frac{e^t(\sin t - \cos t)}{2} + C.

Substituting back $t = \ln x$ :

\int \sin(\ln x)\,dx = \frac{x\bigl(\sin(\ln x) - \cos(\ln x)\bigr)}{2} + C.

Exercise 14: Reduction Formula for Powers of Sine

Hard

A reduction formula expresses $\int \sin^n x\,dx$ in terms of $\int \sin^{n-2}x\,dx$ , lowering the power step by step.

Use integration by parts to derive the reduction formula: $\int \sin^n x\,dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x\,dx.$
Apply the formula to compute $\displaystyle \int_0^{\pi/2} \sin^4 x\,dx$ .

Hint

For question 1, write

\sin^n x = \sin^{n-1}x \cdot \sin x

. Let

u = \sin^{n-1}x

and

dv = \sin x\,dx

. After applying the formula, use the Pythagorean identity

\cos^2 x = 1 - \sin^2 x

to express the resulting integral back in terms of

\int \sin^n x\,dx

and

\int \sin^{n-2}x\,dx

. For question 2, apply the formula twice starting with

n = 4

View Solution

Solution to question 1 (derivation):

Let $u = \sin^{n-1}x$ , $dv = \sin x\,dx \Rightarrow du = (n-1)\sin^{n-2}x\cos x\,dx$ , $v = -\cos x$ .

\int \sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2 x\,dx.

Replace $\cos^2 x = 1 - \sin^2 x$ :

= -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx - (n-1)\int \sin^n x\,dx.

Move the last term to the left side:

n\int \sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx,

\therefore \int \sin^n x\,dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x\,dx. \quad \checkmark

Solution to question 2:

Apply with $n = 4$ on $[0, \pi/2]$ . At both endpoints $\sin^{n-1}x\cos x$ vanishes, so the boundary term is always 0 on this interval.

\int_0^{\pi/2}\sin^4 x\,dx = \frac{3}{4}\int_0^{\pi/2}\sin^2 x\,dx.

Apply with $n = 2$ :

\int_0^{\pi/2}\sin^2 x\,dx = \frac{1}{2}\int_0^{\pi/2}1\,dx = \frac{\pi}{4}.

Therefore:

\int_0^{\pi/2}\sin^4 x\,dx = \frac{3}{4}\cdot\frac{\pi}{4} = \frac{3\pi}{16}.

Exercise 15: Integral Requiring Multiple Combined Strategies

Hard

This final exercise challenges you to chain together substitution, integration by parts, and algebraic manipulation.

Compute $\displaystyle \int x^3\sqrt{x^2+1}\,dx$ using integration by parts (try $u = x^2$ , $dv = x\sqrt{x^2+1}\,dx$ ).
Compute $\displaystyle \int e^{\sqrt[3]{x}}\,dx$ .

Hint

For question 1, integrate

dv = x\sqrt{x^2+1}\,dx

by substituting

t = x^2 + 1

; the result gives

v = \tfrac{(x^2+1)^{3/2}}{3}

. For question 2, let

t = x^{1/3}

to reduce the integral to

3\int t^2 e^t\,dt

, then use the tabular method.

View Solution

Solution to question 1:

Let $u = x^2$ , $dv = x\sqrt{x^2+1}\,dx$ . To find $v$ : set $s = x^2+1$ , $ds = 2x\,dx$ , so

v = \int x\sqrt{x^2+1}\,dx = \frac{(x^2+1)^{3/2}}{3}.

Also $du = 2x\,dx$ . Applying integration by parts:

\int x^3\sqrt{x^2+1}\,dx = \frac{x^2(x^2+1)^{3/2}}{3} - \frac{2}{3}\int x(x^2+1)^{3/2}\,dx.

For the remaining integral, use $s = x^2+1$ again:

\int x(x^2+1)^{3/2}\,dx = \frac{(x^2+1)^{5/2}}{5}.

Therefore:

\int x^3\sqrt{x^2+1}\,dx = \frac{x^2(x^2+1)^{3/2}}{3} - \frac{2(x^2+1)^{5/2}}{15} + C.

Solution to question 2:

Let $t = x^{1/3}$ , so $x = t^3$ and $dx = 3t^2\,dt$ .

\int e^{\sqrt[3]{x}}\,dx = 3\int t^2 e^t\,dt.

From Exercise 8, question 1 (with base function $e^t$ ): $\int t^2 e^t\,dt = e^t(t^2 - 2t + 2)$ . Hence:

\int e^{\sqrt[3]{x}}\,dx = 3e^t(t^2 - 2t + 2) + C.

Substituting back $t = x^{1/3}$ :

\int e^{\sqrt[3]{x}}\,dx = 3e^{\sqrt[3]{x}}\!\left(x^{2/3} - 2x^{1/3} + 2\right) + C.