Question

Integrate 5sinx+3cosx/(sinx+cosx) dx. Limits from 0 to pi/2

Answer 1

Answer:  The answer is $\dfrac{3}{2}\pi.$

Step-by-step explanation:  We are given to integrate a function involving sine and cosine of a variable 'x'.

The integration is as follows -

$I\\\\=\int_{0}^\frac{\pi}{2}\dfrac{5\sin x+3\cos x}{\sin x+\cos x}dx\\\\\\=\int_0^\frac{\pi}{2}\dfrac{3+5\tan x}{1+\tan x}dx\\\\\\=\int_0^\frac{\pi}{2}\dfrac{5(1+\tan x)-2}{1+\tan x}dx\\\\\\=\int_0^\frac{\pi}{2}\left(5-\dfrac{2}{1+\tan x}\right)dx\\\\\\=5\int_0^\frac{\pi}{2}dx-2\int_0^\frac{\pi}{2}\frac{dx}{1+\tan x}\\\\\\=5[x]_0^\frac{\pi}{2}-2[\dfrac{x}{2}+\dfrac{\ln(\cos x+\sin x)}{2}]_0^\frac{\pi}{2}+c\\\\\\=5(\dfrac{\pi}{2}-0)-[x+\ln(\cos x+\sin x)]_0^\frac{\pi}{2}+c\\\\\\=\dfrac{5\pi}{2}-[(\dfrac{\pi}{2}+0)-(0+0)]]=\dfrac{5\pi}{2}-\dfrac{\pi}{2}=2\pi.$

Thus, the answer is $2\pi.$