Integration: Miscellaneous Exercise 7 Questions 20, 24, 25, 27

This lesson works through four integration problems from Miscellaneous Exercise 7: Questions 20, 24, 25, and 27. Two of them rely on the result that $\int e^x\big(f(x)+f'(x)\big)\,dx = e^x f(x) + C$, and the other two use substitution.

Question 20

Find

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

Use $\sin 2x = 2\sin x\cos x$ and $1 + \cos 2x = 2\cos^2 x$:

$\frac{2 + 2\sin x\cos x}{2\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\sin x\cos x}{\cos^2 x} = \sec^2 x + \tan x.$

So the integral becomes

$I = \int e^x\big(\tan x + \sec^2 x\big)\, dx.$

This is of the form $\int e^x\big(f(x) + f'(x)\big)\,dx$ with $f(x) = \tan x$ and $f'(x) = \sec^2 x$. Therefore

$I = e^x \tan x + C.$

Question 24

Evaluate

$I = \int_{\pi/2}^{\pi} e^x\, \frac{1 - \sin x}{1 - \cos x}\, dx.$

Write the trig parts with half angles: $1 - \sin x = 1 - 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ and $1 - \cos x = 2\sin^2\tfrac{x}{2}$. Then

$\frac{1 - \sin x}{1 - \cos x} = \frac{1}{2}\csc^2\tfrac{x}{2} - \cot\tfrac{x}{2}.$

Take $f(x) = -\cot\tfrac{x}{2}$, so $f'(x) = \tfrac{1}{2}\csc^2\tfrac{x}{2}$, and the integrand is $e^x\big(f(x) + f'(x)\big)$. Hence

$I = \Big[-e^x \cot\tfrac{x}{2}\Big]_{\pi/2}^{\pi}.$

At $x = \pi$, $\cot\tfrac{\pi}{2} = 0$; at $x = \tfrac{\pi}{2}$, $\cot\tfrac{\pi}{4} = 1$. So

$I = -\big(0\big) + e^{\pi/2}\cdot 1 = e^{\pi/2}.$

Question 25

Evaluate

$I = \int_{0}^{\pi/4} \frac{\sin x\cos x}{\sin^4 x + \cos^4 x}\, dx.$

Put $t = \sin^2 x$, so $dt = 2\sin x\cos x\, dx$, giving $\sin x\cos x\, dx = \tfrac{1}{2}\, dt$. The limits change: $x = 0 \Rightarrow t = 0$, and $x = \tfrac{\pi}{4} \Rightarrow t = \tfrac{1}{2}$. Also $\cos^4 x = (1 - \sin^2 x)^2 = (1 - t)^2$, so the denominator is $t^2 + (1 - t)^2 = 2t^2 - 2t + 1$. Then

$I = \frac{1}{2}\int_{0}^{1/2} \frac{dt}{2t^2 - 2t + 1}.$

Completing the square

Factor out the $2$ and complete the square:

$2t^2 - 2t + 1 = 2\left(t^2 - t + \tfrac{1}{2}\right) = 2\left[\left(t - \tfrac{1}{2}\right)^2 + \tfrac{1}{4}\right].$

So

$I = \frac{1}{4}\int_{0}^{1/2} \frac{dt}{\left(t - \tfrac{1}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2}.$

Using $\int \frac{dt}{t^2 + a^2} = \tfrac{1}{a}\tan^{-1}\tfrac{t}{a}$ with $a = \tfrac{1}{2}$:

$I = \frac{1}{4}\cdot 2\,\Big[\tan^{-1}(2t - 1)\Big]_{0}^{1/2} = \frac{1}{2}\Big[\tan^{-1}(2t - 1)\Big]_{0}^{1/2}.$

Evaluating: at $t = \tfrac{1}{2}$ we get $\tan^{-1}(0)=0$; at $t = 0$ we get $\tan^{-1}(-1) = -\tfrac{\pi}{4}$. So

$I = \frac{1}{2}\left(0 + \frac{\pi}{4}\right) = \frac{\pi}{8}.$

Question 27

Evaluate

$I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}}\, dx.$

Put $t = \sin x - \cos x$, so $dt = (\cos x + \sin x)\, dx$, which matches the numerator. Squaring, $t^2 = \sin^2 x + \cos^2 x - 2\sin x\cos x = 1 - \sin 2x$, so $\sin 2x = 1 - t^2$. Change the limits:

$x = \tfrac{\pi}{6} \Rightarrow t = \sin\tfrac{\pi}{6} - \cos\tfrac{\pi}{6} = \frac{1 - \sqrt{3}}{2},$

$x = \tfrac{\pi}{3} \Rightarrow t = \sin\tfrac{\pi}{3} - \cos\tfrac{\pi}{3} = \frac{\sqrt{3} - 1}{2}.$

The integral becomes

$I = \int_{(1-\sqrt{3})/2}^{(\sqrt{3}-1)/2} \frac{dt}{\sqrt{1 - t^2}} = \Big[\sin^{-1} t\Big]_{(1-\sqrt{3})/2}^{(\sqrt{3}-1)/2}.$

Using $\sin^{-1}(-x) = -\sin^{-1} x$, the two endpoints are negatives of each other, so

$I = \sin^{-1}\frac{\sqrt{3} - 1}{2} - \sin^{-1}\frac{1 - \sqrt{3}}{2} = 2\sin^{-1}\frac{\sqrt{3} - 1}{2}.$

Key takeaways

• If an integrand can be written as $e^x\big(f(x) + f'(x)\big)$, the integral is simply $e^x f(x) + C$.
• Double-angle and half-angle identities are the usual way to reshape trig fractions into that form.
• For definite integrals, substitute, change the limits to the new variable, and complete the square when you need an inverse tangent or inverse sine standard form.