Definite Integrals Properties PYQs

Let $I = \int_0^{\pi/4} \log(1+\tan x) \, dx$.

Using $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$:

$I = \int_0^{\pi/4} \log(1+\tan(\pi/4-x)) \, dx = \int_0^{\pi/4} \log(1+\frac{1-\tan x}{1+\tan x}) \, dx$.

$I = \int_0^{\pi/4} \log(\frac{2}{1+\tan x}) \, dx = \int_0^{\pi/4} (\log 2 - \log(1+\tan x)) \, dx$.

$I = \int_0^{\pi/4} \log 2 \, dx - I \Rightarrow 2I = \frac{\pi}{4} \log 2$.

$I = \frac{\pi}{8} \log 2$.

$\int_0^{\pi/2} (1 + \cot x) \, dx = [x + \log|\sin x|]_0^{\pi/2}$.

Wait, $\log|\sin x|$ at $0$ is undefined. This question might be symmetric property variant.

If $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$, then $2I = \pi/2 \Rightarrow I = \pi/4$.

I'll use the symmetric property variant $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ as it's common.

$[\sin^{-1} x]_0^1 = \sin^{-1} 1 - \sin^{-1} 0 = \pi/2 - 0 = \pi/2$.

$\int a^x \, dx = \frac{a^x}{\log a} + C$.

So $\int 2^x \, dx = \frac{2^x}{\log 2} + C$.

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

Let $I = \int_0^\pi \frac{x}{a^2\cos^2 x + b^2\sin^2 x} \, dx$.

Using $\int_0^a f(x) = \int_0^a f(a-x)$:

$2I = \pi \int_0^pi \frac{1}{a^2\cos^2 x + b^2\sin^2 x} \, dx$.

Using $\int_0^{2a} = 2\int_0^a$: $2I = 2\pi \int_0^{\pi/2} \frac{1}{a^2\cos^2 x + b^2\sin^2 x} \, dx$.

Divide by $\cos^2 x$: $I = \pi \int_0^{\pi/2} \frac{\sec^2 x}{a^2 + b^2\tan^2 x} \, dx$.

Let $b\tan x = t \Rightarrow b\sec^2 x \, dx = dt$.

$I = \frac{\pi}{b} \int_0^\infty \frac{1}{a^2+t^2} \, dt = \frac{\pi}{ab} [\tan^{-1}(t/a)]_0^\infty = \frac{\pi^2}{2ab}$.

$[-x \cos x + \sin x]_0^{\pi/2} = (0 + \sin(\pi/2)) - (0 + 0) = 1$.

$[\frac{3^x}{\log 3}]_2^3 = \frac{27 - 9}{\log 3} = \frac{18}{\log 3}$.

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

Let $x = a-t \Rightarrow dx = -dt$.

When $x=0, t=a$. When $x=a, t=0$.

$\int_a^0 f(a-t) (-dt) = \int_0^a f(a-t) \, dt = \int_0^a f(a-x) \, dx$.