Question

Find the integrals of the function: $sin^2(2x+5)$

Answer 1

Answer:

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

Step-by-step explanation:

Formula used:

$cos2A=1-2sin^2A\\\\sin^2A=\frac{1-cos2A}{2}$

Now,

$\int{sin^2(2x+5)}\:dx\\\\=\int{\frac{1-cos2(2x+5)}{2}}\:dx\\\\=\frac{1}{2}\int{[1-cos(2x+10)]}\:dx\\\\=\frac{1}{2}[x-\frac{sin(2x+10)}{2}]+c\\\\=\frac{x}{2}-\frac{sin(2x+10)}{4}+c$