1 Megul

## Integration Of Powers Of Sine And Cosine Homework Answers

Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach.

Example 8.2.1 Evaluate $\ds\int \sin^5 x\,dx$. Rewrite the function: $$\int \sin^5 x\,dx=\int \sin x \sin^4 x\,dx= \int \sin x (\sin^2 x)^2\,dx= \int \sin x (1-\cos^2 x)^2\,dx.$$ Now use $u=\cos x$, $du=-\sin x\,dx$: \eqalign{ \int \sin x (1-\cos^2 x)^2\,dx&=\int -(1-u^2)^2\,du\cr &=\int -(1-2u^2+u^4)\,du\cr &=-u+{2\over3}u^3-{1\over5}u^5+C\cr &=-\cos x+{2\over3}\cos^3 x-{1\over5}\cos^5x+C.\cr }

Example 8.2.2 Evaluate $\ds\int \sin^6 x\,dx$. Use $\ds \sin^2x =(1-\cos(2x))/2$ to rewrite the function: \eqalign{ \int \sin^6 x\,dx=\int (\sin^2 x)^3\,dx&= \int {(1-\cos 2x)^3\over 8}\,dx\cr &={1\over 8}\int 1-3\cos 2x+3\cos^2 2x-\cos^3 2x\,dx.\cr} Now we have four integrals to evaluate: $$\int 1\,dx=x$$ and $$\int -3\cos 2x\,dx = -{3\over 2}\sin 2x$$ are easy. The $\ds \cos^3 2x$ integral is like the previous example: \eqalign{ \int -\cos^3 2x\,dx&=\int -\cos 2x\cos^2 2x\,dx\cr &=\int -\cos 2x(1-\sin^2 2x)\,dx\cr &=\int -{1\over 2}(1-u^2)\,du\cr &=-{1\over 2}\left(u-{u^3\over 3}\right)\cr &=-{1\over 2}\left(\sin 2x-{\sin^3 2x\over 3}\right).} And finally we use another trigonometric identity, $\ds \cos^2x=(1+\cos(2x))/2$: $$\int 3\cos^2 2x\,dx=3\int {1+\cos 4x\over 2}\,dx= {3\over 2}\left(x+{\sin 4x\over 4}\right).$$ So at long last we get $$\int \sin^6 x\,dx = {x\over8} -{3\over 16}\sin 2x -{1\over 16}\left(\sin 2x-{\sin^3 2x\over 3}\right) +{3\over 16}\left(x+{\sin 4x\over 4}\right)+C.$$

Example 8.2.3 Evaluate $\ds\int\! \sin^2x\cos^2x\,dx$. Use the formulas $\ds \sin^2x =(1-\cos(2x))/2$ and $\ds \cos^2x =(1+\cos(2x))/2$ to get: $$\int \sin^2x\cos^2x\,dx=\int {1-\cos(2x)\over2}\cdot {1+\cos(2x)\over2}\,dx.$$ The remainder is left as an exercise.

## Exercises 8.2

Find the antiderivatives.

Ex 8.2.1 $\ds\int \sin^2 x\,dx$ (answer)

Ex 8.2.2 $\ds\int \sin^3 x\,dx$ (answer)

Ex 8.2.3 $\ds\int \sin^4 x\,dx$ (answer)

Ex 8.2.4 $\ds\int \cos^2 x\sin^3 x\,dx$ (answer)

Ex 8.2.5 $\ds\int \cos^3 x\,dx$ (answer)

Ex 8.2.6 $\ds\int \sin^2 x\cos^2 x\,dx$ (answer)

Ex 8.2.7 $\ds\int \cos^3 x \sin^2 x\,dx$ (answer)

Ex 8.2.8 $\ds\int \sin x (\cos x)^{3/2}\,dx$ (answer)

Ex 8.2.9 $\ds\int \sec^2 x\csc^2 x\,dx$ (answer)

Ex 8.2.10 $\ds\int \tan^3x \sec x\,dx$ (answer)

There are several approaches. One, that has been given in previous answers, uses the fact that $\cos(\theta)=\frac12(e^{i\theta}+e^{i\theta})$ and that $\int_0^{2\pi}e^{in\theta}\,\mathrm{d}\theta=2\pi$ if $n=0$, and vanishes otherwise, to get \begin{align} \int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta &=\frac14\int_0^{2\pi}\cos^{2n}(\theta)\,\mathrm{d}\theta\\ &=\frac1{4^{n+1}}\int_0^{2\pi}\left(e^{i\theta}+e^{i\theta}\right)^{2n}\,\mathrm{d}\theta\\ &=\frac{2\pi}{4^{n+1}}\binom{2n}{n} \end{align}

Another approach is to integrate by parts \begin{align} \int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta &=\int_0^{\pi/2}\cos^{2n-1}(\theta)\,\mathrm{d}\sin(\theta)\\ &=(2n-1)\int_0^{\pi/2}\sin^2(\theta)\cos^{2n-2}(\theta)\,\mathrm{d}\theta\\ &=(2n-1)\int_0^{\pi/2}\left(\cos^{2n-2}(\theta)-\cos^{2n}(\theta)\right)\,\mathrm{d}\theta\\ &=\frac{2n-1}{2n}\int_0^{\pi/2}\cos^{2n-2}(\theta)\,\mathrm{d}\theta\\ \end{align} and use induction to get $$\int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta =\frac\pi2\prod_{k=1}^n\frac{2k-1}{2k}$$

Note that \begin{align} \frac\pi2\prod_{k=1}^n\frac{2k-1}{2k} &=\frac\pi2\frac{(2n)!}{(2^nn!)^2}\\ &=\frac{\pi}{2^{2n+1}}\binom{2n}{n} \end{align}

answered Jun 23 '15 at 23:19