Resolved: "Evaluate the following integrals. ∫sin2xcos4xdx"

tabita57i

Answered question

2021-10-05

Evaluate the following integrals.

\int \sin^{2} x \cos^{4} x d x

2k1enyvp

Skilled2021-10-06Added 94 answers

To Determine: find the integral

\int \sin^{2} x \cos^{4} x d x

Given: we have an integral

\int \sin^{2} x \cos^{4} x d x

Explanation: we will solve the integral as follows
Put

u = \cos x

then

d u = - \sin x d x

also

\frac{d u}{- \sin x} = d x

Also

- \frac{d u}{\sqrt{1 - \cos^{2} x}} = d x

The integral will also be written as

\int \sin^{x} \cos^{4} x d x = \int (1 - \cos^{2} x) \cos^{4} x d x

= \int \frac{(1 - \cos^{2} x)}{\sqrt{1 - \cos^{2} x}} \cos^{4} x d x

Now putting the values we have

= - \int u^{4} \sqrt{1 - u^{2}} d u

Again putting

u = \sin t

then

d u = \cos t d t

Integral becomes

= - \int \sin^{4} (t) \cos (t) \cos t d t

= - \int \sin^{4} (t) \cos^{2} (t) d t

= - \int \sin^{4} (t) - \sin^{6} (t) d t

= - (\int \sin^{4} t d t - \int \sin^{6} t d t)

We know that

\int \sin^{n} x d x = - \frac{\cos x \sin^{n - 1} x}{n} + \frac{n - 1}{n} \int \sin^{n - 2} x d x

\int \sin^{4} t d v = - \frac{1}{4} \sin^{3} t \cos t + \frac{3}{8} (t - \frac{1}{2} \sin (2 t))

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