Question

integration of sin square x cos square x​

Answer 1

Answer:

Step-by-step explanation:

$Sin^{2} x. Cos^2x.dx$

Sin^2x . 1-Sin^2x . dx

sin^x and -sin^2x gets cancelled

1 . dx

=> dx

Answer 2

Answer:

$\displaystyle\int\sin^2x\,\cos^2x\,dx\\=\tfrac14\int(2\sin x\cos x)^2\,dx\\=\tfrac14\int\sin^22x\,dx=\tfrac18\int(1-\cos4x)\,dx\\\\=\tfrac18(x-\tfrac14\sin4x)+C\\\\=\tfrac18x-\tfrac1{32}\sin4x+C$