Question

INTEGRATE...............

Answer 1

Answer:

Hey mate,

ur ans is

$\frac{-cos^3}{3}+\frac{cos^5x}{5}$

Step-by-step explanation:

let cos x =t

sin dx =-dt

∫sin²xcos²x.sin x dx

-∫sin²x(1-sin²x)dt

-∫t²(1-t²)dt

-∫t²-t∧4 dt

-t³/3+t^5/5 +c

subsituting  the value of t in ↑↑↑

we get $\frac{-cos^3}{3}+\frac{cos^5x}{5}$

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Answer 2

Answer:

Hey mate,

ur ans is

\begin{lgathered}\frac{-cos^3}{3}+\frac{cos^5x}{5} \\\end{lgathered}

3

−cos

3

+

5

cos

5

x

Step-by-step explanation:

let cos x =t

sin dx =-dt

∫sin²xcos²x.sin x dx

-∫sin²x(1-sin²x)dt

-∫t²(1-t²)dt

-∫t²-t∧4 dt

-t³/3+t^5/5 +c

subsituting the value of t in ↑↑↑

we get \begin{lgathered}\frac{-cos^3}{3}+\frac{cos^5x}{5} \\\end{lgathered}

3

−cos

3

+

5

cos

5

x