"Evaluate the indefinite integral. ∫9sin5(x)cos3(x)dx" was answered by our community

Barbara Schroder

Answered question

2021-11-10

Evaluate the indefinite integral.

\int 9 \sin^{5} (x) \cos^{3} (x) d x

Melinda Olson

Beginner2021-11-11Added 20 answers

Step 1: Given that:

\int 9 \sin^{5} (x) \cos^{3} (x) d x

We need to Evaluate the integral
Step 2: Solving the integral

\int 9 \sin^{5} (x) \cos^{3} (x) d x = \int 9 \sin^{5} (x) (1 - \sin^{2} (x)) \cos (x) d x

Substituting

\sin x = t

\cos x d x = d t

d x = \frac{d t}{\cos x}

Plugging all the values into the integral we get,

\int 9 \sin^{5} x (1 - \sin^{2} x) \cos x d x = \int 9 t^{5} (1 - t^{2}) \cos x \frac{d t}{\cos x}

= \int 9 t^{5} (1 - t^{2}) d t

= 9 \int t^{5} d t - 9 \int t^{7} d t

= 9 (\frac{t^{6}}{6}) - 9 (\frac{t^{8}}{8}) + C

= \frac{3 \sin^{6} (x)}{2} - \frac{9 \sin^{8} (x)}{8} + C

Therefore,

\int 9 \sin^{5} (x) \cos^{3} (x) d x = \frac{\sin^{6} (x) (12 - 9 \sin^{2} (x))}{8} + C

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