Question

Evaluate the following indefinite integrals.

Answer 1

sin^4 x = sin^2x * sin^2x = (1-cos^2x)(sin^2x)
see if it helps ?

Answer 2

if we adopt shubham strategy,and put cosx=t,we get..
\[-32\int\limits_{}^{}(1-t ^{2})^{3/2}dt\]

Answer 3

it will be better if we use the sine half angle...

Answer 4

\[32\int\limits_{}^{}(\sin ^{4}x) dx\]
32\[\int\limits_{}^{}(((1-\cos(2x))/2)^{2}dx\]
\[8 \int\limits_{}^{}(1-2\cos(2x)+\cos ^{2}(2x))dx\]
\[8\int\limits_{}^{}dx-8\int\limits_{}^{}2 \cos(2x)dx+8\int\limits_{}^{}(1+\cos(4x))/2\]

Answer 5

i hope u can solve it further....

Answer 6

yes or no @Albertoimus

Answer 7

i kinda did it a different way... imma try it this way.

Answer 8

what would be the answer after simplifying?

Answer 9

8x-16(sin(2x))/2)+8(x/2)+4(sin(4x)/4)
8x-8 sin(2x)+4x+sin(4x)