Question

sin²(2x + 5).dx
Integrate the function

Answer 1

HELLOD EAR,

Given function is intgral of sin²(2x+5).dx

NOW, $\sf{\Rightarrow\int{\frac{1-cos(4x+ 10)}{2}}\,dx}$

$\sf{\Rightarrow1/2[\int{1}\,dx-\int{cos(4x+ 10)}\,dx]}$

$\sf{\Rightarrow x/2 - 1/2(\frac{sin(4x+10)}{4})}$

$\sf{\Rightarrow \frac{1}{2}x-\frac{1}{8}sin(5x+10)+ C}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Solution-

Given,

$f(x)=\sin^2 (2x+5)$

We know that,

$\cos 2x=1-2sin^2 x$

$sin^2 x=\frac{1-\cos 2x}{2}$

Now,

$\int f(x).dx$

$=\int \sin^2 (2x+5).dx$

$=\int \frac{1-\cos(4x+10)}{2}.dx$

$=\frac{1}{2}[\int dx-\int \cos(4x+10).dx]$

$=\frac{1}{2}[x-\frac{1}{4}\sin(4x+10)+c]$

$=\frac{1}{2}x-\frac{1}{8}\sin(4x+10)+c$