Integration by Substitution: Trigonometric Questions (Exercise 7.2)

This lesson solves the trigonometric integrals of Exercise 7.2, every one of them by the substitution method. In each case we pick an inner expression, call it $t$, replace it along with its derivative, and reduce the integral to a standard result.

Powers of tangent

Evaluate

$\int \frac{\tan^{4}(\sqrt{x})\,\sec^{2}(\sqrt{x})}{\sqrt{x}}\,dx.$

Put $\tan(\sqrt{x}) = t$. Then

$\frac{d}{dx}\,\tan(\sqrt{x}) = \sec^{2}(\sqrt{x})\cdot\frac{1}{2\sqrt{x}},$

so $\dfrac{\sec^{2}(\sqrt{x})}{\sqrt{x}}\,dx = 2\,dt$. The integral becomes

$\int t^{4}\,(2\,dt) = \frac{2t^{5}}{5} = \frac{2}{5}\,\tan^{5}(\sqrt{x}) + c.$

Sine of an arctangent

Evaluate

$\int \frac{\sin(\tan^{-1}x)}{1+x^{2}}\,dx.$

Put $\tan^{-1}x = t$, so $\dfrac{1}{1+x^{2}}\,dx = dt$. Then

$\int \sin t\,dt = -\cos t = -\cos(\tan^{-1}x) + c.$

Odd power of sine times cosine squared

Evaluate

$\int \sin^{3}x\,\cos^{2}x\,dx = \int \sin^{2}x\,\cos^{2}x\,\sin x\,dx.$

Write $\sin^{2}x = 1-\cos^{2}x$ and put $\cos x = t$, so $\sin x\,dx = -dt$:

$\int (1-t^{2})\,t^{2}\,(-dt) = -\int (t^{2}-t^{4})\,dt = -\frac{t^{3}}{3}+\frac{t^{5}}{5}.$

Substituting back,

$-\frac{\cos^{3}x}{3}+\frac{\cos^{5}x}{5} + c.$

Sine of x over sine of x plus a

Evaluate

$\int \frac{\sin x}{\sin(x+a)}\,dx.$

Put $x+a = t$, so $x = t-a$ and $dx = dt$:

$\int \frac{\sin(t-a)}{\sin t}\,dt.$

Expand $\sin(t-a) = \sin t\cos a - \cos t\sin a$ and split:

$\int \Big(\cos a - \sin a\,\cot t\Big)\,dt = t\cos a - \sin a\,\log|\sin t|.$

Since $\int \cot t\,dt = \log|\sin t|$, substituting back gives

$x\cos a - \sin a\,\log|\sin(x+a)| + c.$

Arcsine over the root

Evaluate

$\int \frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\,dx.$

Since $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^{2}}}$, put $\sin^{-1}x = t$, so $\dfrac{1}{\sqrt{1-x^{2}}}\,dx = dt$:

$\int t\,dt = \frac{t^{2}}{2} = \frac{(\sin^{-1}x)^{2}}{2} + c.$

A quotient of sine and cosine terms

Evaluate

$\int \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x}\,dx.$

Take $2$ out of the denominator:

$\frac{1}{2}\int \frac{2\cos x - 3\sin x}{3\cos x + 2\sin x}\,dx.$

Put $3\cos x + 2\sin x = t$. Then $dt = (-3\sin x + 2\cos x)\,dx = (2\cos x - 3\sin x)\,dx$, which is exactly the numerator:

$\frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\log|3\cos x + 2\sin x| + c.$

Cosine of a root

Evaluate

$\int \frac{\cos(\sqrt{x})}{\sqrt{x}}\,dx.$

Put $\sqrt{x} = t$, so $\dfrac{1}{2\sqrt{x}}\,dx = dt$, that is $\dfrac{1}{\sqrt{x}}\,dx = 2\,dt$:

$\int \cos t\,(2\,dt) = 2\sin t = 2\sin(\sqrt{x}) + c.$

Secant squared over a squared bracket

Evaluate

$\int \frac{\sec^{2}x}{(1-\tan x)^{2}}\,dx.$

Put $1-\tan x = t$, so $-\sec^{2}x\,dx = dt$, that is $\sec^{2}x\,dx = -dt$:

$\int \frac{-dt}{t^{2}} = \frac{1}{t} = \frac{1}{1-\tan x} + c.$

Root of tangent over sine cosine

Evaluate

$\int \frac{\sqrt{\tan x}}{\sin x\,\cos x}\,dx.$

Multiply and divide by $\cos x$:

$\int \frac{\sqrt{\tan x}}{\tan x\,\cos^{2}x}\,dx = \int \frac{\sec^{2}x}{\sqrt{\tan x}}\,dx.$

Put $\tan x = t$, so $\sec^{2}x\,dx = dt$:

$\int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{\tan x} + c.$

An eighth power in the denominator

Evaluate

$\int \frac{x^{3}\,\sin\!\big(\tan^{-1}(x^{4})\big)}{1+x^{8}}\,dx.$

Put $\tan^{-1}(x^{4}) = t$. Then

$dt = \frac{1}{1+(x^{4})^{2}}\cdot 4x^{3}\,dx,$

so $\dfrac{x^{3}}{1+x^{8}}\,dx = \dfrac{1}{4}\,dt$. The integral becomes

$\int \sin t\cdot\frac{1}{4}\,dt = -\frac{1}{4}\cos t = -\frac{1}{4}\cos\!\big(\tan^{-1}(x^{4})\big) + c.$

One over sine squared cosine squared

Evaluate

$\int \frac{dx}{\sin^{2}x\,\cos^{2}x}.$

Write the numerator $1 = \sin^{2}x + \cos^{2}x$ and split:

$\int \frac{\sin^{2}x + \cos^{2}x}{\sin^{2}x\,\cos^{2}x}\,dx = \int \frac{1}{\cos^{2}x}\,dx + \int \frac{1}{\sin^{2}x}\,dx.$

These are standard integrals:

$\int \sec^{2}x\,dx + \int \csc^{2}x\,dx = \tan x - \cot x + c.$

Key takeaways

• The whole method is choosing an inner function $t$ so that its derivative already sits in the integrand, then replacing both.
• Identities like $\sin^{2}x = 1-\cos^{2}x$ and $1 = \sin^{2}x+\cos^{2}x$ turn awkward integrands into ones a substitution can handle.
• After integrating in $t$, always substitute back to express the answer in $x$ and add the constant $c$.