Get the answer to "Evaluate the integral. ∫cos2xsin⁡(2x)dx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

prsategazd

Answered question

2021-12-11

Evaluate the integral.

\int \cos^{2} x \sin (2 x) d x

Answer & Explanation

temzej9

Beginner2021-12-12Added 30 answers

Step 1
Given integral to evaluate
$\int \cos^{2} x \sin (2 x) d x$
substitute $\sin (2 x) = 2 \sin x \cos x$
$\int \cos^{2} x \sin (2 x) d x = \int \cos^{2} x 2 \sin x \cos x d x$
$= \int 2 \cos^{3} x \sin x d x$
Step 2
Now put
$t = \cos x$
$d t = - \sin x d x$
so the integral becomes
$\int - 2 t^{3} d t = - 2 \int t^{3} d t$
$= - 2 [\frac{t^{4}}{4}] + c$
$= - \frac{t^{4}}{2} + c$
put back value of t.
$\int \cos^{2} x \sin (2 x) d x = - \frac{\cos^{4} x}{2} + c$

ambarakaq8

Beginner2021-12-13Added 31 answers

\int \cos^{2} (x) \sin (2 x) d x

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

= - 2 \int u^{3} d u

Now we calculate:

\int u^{3} d u

= \frac{u^{4}}{4}

We substitute the already calculated integrals:

- 2 \int u^{3} d u

= - \frac{u^{4}}{2}

Reverse replacement

u = \cos (x) :

= - \frac{\cos^{4} (x)}{2}

Answer:

= - \frac{\cos^{4} (x)}{2} + C

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