Lesson notes

This lesson works through several definite integrals of trigonometric functions from Exercise 7.8. For each one we find the antiderivative using a standard result, then substitute the upper and lower limits.

Question 4

Evaluate

$\int_0^{\pi/4} \sin 2x \, dx.$

The antiderivative of $\sin 2x$ is $-\tfrac{1}{2}\cos 2x$, so

$\left[-\tfrac{1}{2}\cos 2x\right]_0^{\pi/4} = -\tfrac{1}{2}\left(\cos\tfrac{\pi}{2} - \cos 0\right) = -\tfrac{1}{2}(0 - 1) = \tfrac{1}{2}.$

Question 7

Evaluate

$\int_0^{\pi/4} \tan x \, dx.$

Using $\int \tan x \, dx = \log|\sec x|$,

$\left[\log|\sec x|\right]_0^{\pi/4} = \log \sec\tfrac{\pi}{4} - \log \sec 0 = \log \sqrt{2} - \log 1.$

Since $\sqrt{2} = 2^{1/2}$, this equals $\tfrac{1}{2}\log 2$.

Question 8

Evaluate

$\int_{\pi/6}^{\pi/4} \operatorname{cosec} x \, dx.$

Using $\int \operatorname{cosec} x \, dx = \log|\operatorname{cosec} x - \cot x|$,

$\left[\log|\operatorname{cosec} x - \cot x|\right]_{\pi/6}^{\pi/4}.$

At $x = \tfrac{\pi}{4}$, $\operatorname{cosec} x = \sqrt{2}$ and $\cot x = 1$. At $x = \tfrac{\pi}{6}$, $\operatorname{cosec} x = 2$ and $\cot x = \sqrt{3}$. So

$\log(\sqrt{2} - 1) - \log(2 - \sqrt{3}) = \log\frac{\sqrt{2} - 1}{2 - \sqrt{3}}.$

Question 9

Evaluate

$\int_0^{1} \frac{dx}{\sqrt{1 - x^2}}.$

This is $\left[\sin^{-1} x\right]_0^{1} = \sin^{-1} 1 - \sin^{-1} 0 = \tfrac{\pi}{2} - 0 = \tfrac{\pi}{2}.$

Question 12

Evaluate

$\int_0^{\pi/2} \cos^2 x \, dx.$

Use the identity $\cos^2 x = \tfrac{1 + \cos 2x}{2}$:

$\tfrac{1}{2}\int_0^{\pi/2} (1 + \cos 2x)\, dx = \tfrac{1}{2}\left[x + \tfrac{\sin 2x}{2}\right]_0^{\pi/2}.$

Since $\sin \pi = 0$ and $\sin 0 = 0$,

$\tfrac{1}{2}\left(\tfrac{\pi}{2} + 0\right) = \tfrac{\pi}{4}.$

Question 17

Evaluate

$\int_0^{\pi/4} \left(2\sec^2 x + x^3 + 2\right) dx.$

Split into three integrals:

$\left[2\tan x\right]_0^{\pi/4} + \left[\tfrac{x^4}{4}\right]_0^{\pi/4} + \left[2x\right]_0^{\pi/4}.$

First term: $2(\tan\tfrac{\pi}{4} - \tan 0) = 2(1 - 0) = 2.$

Second term: $\tfrac{1}{4}\left(\tfrac{\pi}{4}\right)^4 = \tfrac{\pi^4}{4 \cdot 256} = \tfrac{\pi^4}{1024}.$

Third term: $2 \cdot \tfrac{\pi}{4} = \tfrac{\pi}{2}.$

Adding these,

$2 + \frac{\pi^4}{1024} + \frac{\pi}{2}.$

Question 18

Evaluate

$\int_0^{\pi} \left(\sin^2 \tfrac{x}{2} - \cos^2 \tfrac{x}{2}\right) dx.$

Take the negative outside and use $\cos^2\theta - \sin^2\theta = \cos 2\theta$ with $\theta = \tfrac{x}{2}$:

$-\int_0^{\pi} \left(\cos^2 \tfrac{x}{2} - \sin^2 \tfrac{x}{2}\right) dx = -\int_0^{\pi} \cos x \, dx = -\left[\sin x\right]_0^{\pi}.$

Since $\sin \pi = 0$ and $\sin 0 = 0$, the value is $0$.

Question 21

Evaluate

$\int_1^{\sqrt{3}} \frac{dx}{1 + x^2}.$

Using $\int \tfrac{dx}{1 + x^2} = \tan^{-1} x$,

$\left[\tan^{-1} x\right]_1^{\sqrt{3}} = \tan^{-1}\sqrt{3} - \tan^{-1} 1 = \tfrac{\pi}{3} - \tfrac{\pi}{4} = \tfrac{4\pi - 3\pi}{12} = \tfrac{\pi}{12}.$

Question 22

Evaluate

$\int_0^{2/3} \frac{dx}{4 + 9x^2}.$

Take $9$ outside so the coefficient of $x^2$ is $1$:

$\tfrac{1}{9}\int_0^{2/3} \frac{dx}{\left(\tfrac{2}{3}\right)^2 + x^2}.$

Using $\int \tfrac{dx}{a^2 + x^2} = \tfrac{1}{a}\tan^{-1}\tfrac{x}{a}$ with $a = \tfrac{2}{3}$,

$\tfrac{1}{9} \cdot \tfrac{3}{2}\left[\tan^{-1}\tfrac{3x}{2}\right]_0^{2/3} = \tfrac{1}{6}\left(\tan^{-1} 1 - \tan^{-1} 0\right) = \tfrac{1}{6} \cdot \tfrac{\pi}{4} = \tfrac{\pi}{24}.$

Key takeaways

• A definite integral is found by computing the antiderivative and then substituting the upper limit minus the lower limit.
• Identities like $\cos^2 x = \tfrac{1 + \cos 2x}{2}$ and $\cos^2\theta - \sin^2\theta = \cos 2\theta$ simplify integrals before evaluating them.
• Integrals of the forms $\tfrac{1}{\sqrt{1 - x^2}}$ and $\tfrac{1}{a^2 + x^2}$ give inverse sine and inverse tangent results.