Solved: "Find the indefinite integral. ∫sin72xcos⁡2xdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

akutagawatj

Answered question

2021-12-07

Find the indefinite integral.

\int \sin^{7} 2 x \cos 2 x d x

Answer & Explanation

Luis Sullivan

Beginner2021-12-08Added 11 answers

Step 1
Think about the provided integral $\int \sin^{7} 2 x \cos 2 x d x$
Step 2
Substitute $\sin^{7} 2 x = t$ then differentiate
$7 \sin^{6} 2 x \cos 2 x (2) d x = d t$
$14 \sin^{6} 2 x \cos 2 x = d t$
Simplify as
$\int \sin^{7} 2 x \cos 2 x d x = \int \sin^{6} 2 x \sin 2 x \cos 2 x d t$
$= \frac{1}{14} \int t^{\frac{1}{7}} d t$
$= \frac{1}{14} (\frac{t^{\frac{1}{7} + 1}}{\frac{1}{7} + 1}) + c$
$= \frac{1}{14} (\frac{t^{\frac{8}{7}}}{\frac{8}{7}}) + c$
$= \frac{1}{16} t^{\frac{8}{7}} + c$
Step 3
Substitute back $\sin^{7} 2 x = t$
$\int \sin^{7} 2 x \cos 2 x d x = \frac{1}{16} (\sin^{8} 2 x) + c$

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