"Evaluate the following integrals. ∫sin2xcos2xdx" was answered by our experts

sanuluy

Answered question

2021-10-13

Evaluate the following integrals.

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

Faiza Fuller

Skilled2021-10-14Added 108 answers

We have to evaluate the integral:

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

Since we know the identity,

2 \sin x \cos x = \sin 2 x

Therefore multiplying and dividing by

2^{2}

\frac{2^{2}}{2^{2}} \int \sin^{2} x \cos^{2} x d x = \frac{1}{4} \int 2^{2} \sin^{2} x \cos^{2} x d x

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

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

= \frac{1}{4} \int \sin^{2} 2 x d x

We have identity,

\cos^{2} x - \sin^{2} x = \cos 2 x

\cos 2 x = 1 - 2 \sin^{2} x (\cos^{2} x + \sin^{2} x = 1)

\sin^{2} x = \frac{1 - \cos 2 x}{2}

Therefore replacing x to 2x, we get

\sin^{2} 2 x = \frac{1 - \cos 4 x}{2} d x

= \frac{1}{8} \int (1 - \cos 4 x) d x

= \frac{1}{8} (\int d x - \int \cos 4 x d x)

= \frac{1}{8} x - \frac{1}{8} (\frac{\sin 4 x}{4}) + C

= \frac{1}{8} (x - \frac{\sin 4 x}{4}) + C

Where, C is an arbitrary constant.
Hence, value of integral is

\frac{1}{8} (x - \frac{\sin 4 x}{4}) + C

Note: we have used following formula,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

\int \cos x d x = \sin x + C

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