Evaluate the following indefinite integrals.

Question

anonymous · Answer

attachment

shubhamsrg · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

i hope u can solve it further....

anonymous · Answer

yes or no @Albertoimus

anonymous · Answer

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

anonymous · Answer

what would be the answer after simplifying?

anonymous · Answer

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