# Evaluate int sin^{27}{(x)}cos^{46}{(x)} dx.

## Question:

Evaluate {eq}\displaystyle \int \sin^{27}{(x)} \cos^{46}{(x)} dx {/eq}.

## Integration:

To evaluate the integral, first, we use the trigonometry relation between {eq}\sin (\theta) \: and \: \cos (\theta) {/eq} i.e

{eq}\sin^2(\theta)+\cos^2(\theta)=1 {/eq}

Then, we use the substitution to solve the integral because the integral becomes the form of {eq}\int h(g(x))g'(x) dx {/eq}

## Answer and Explanation:

We have to evaluate {eq}\displaystyle \int \sin^{27}{(x)} \cos^{46}{(x)} dx {/eq}.

{eq}\begin{align} \int \sin ^{27}\left(x\right)\cos ^{46}\left(x\right)dx &=\int \sin ^{26}\left(x\right)\sin \left(x\right)\cos ^{46}\left(x\right)dx\\ &=\int \left(\sin ^2\left(x\right)\right)^{13}\sin \left(x\right)\cos ^{46}\left(x\right)dx\\ &=\int \left(1-\cos ^2\left(x\right)\right)^{13}\sin \left(x\right)\cos ^{46}\left(x\right)dx\ & \left [ \because \sin ^2\left(x\right)=1-\cos ^2\left(x\right) \right ]\\ \end{align} {/eq}

We use the substitution {eq}\cos (x)=u {/eq}

Differentiate both sides with respect to {eq}x {/eq}

{eq}\begin{align} \cos (x) &=u\\ \dfrac{\mathrm{d} }{\mathrm{d} x}(\cos x) &=\dfrac{\mathrm{d} u}{\mathrm{d} x}\\ -\sin (x)dx &=du \end{align} {/eq}

Applying this value, we have:

{eq}\begin{align} \int \left(1-\cos ^2\left(x\right)\right)^{13}\sin \left(x\right)\cos ^{46}\left(x\right)dx &=\int \:-u^{46}\left(1-u^2\right)^{13}du\\ &=\int -u^{46}\left ( 1-13u^2+78u^4-286u^6+715u^8-1287u^{10}+1716u^{12}-1716u^{14}+1287u^{16}-715u^{18}+286u^{20}-78u^{22}+13u^{24}-u^{26} \right )du\ \left[ \text{Applying the bionomial theorem} \right]\\ &=\int \left (-u^{46}+13u^{48}-78u^{50}+286u^{52}-715u^{54}+1287u^{56}-1716u^{58}+1716u^{60}-1287u^{62}+715u^{64}-286u^{66}+78u^{68}-13u^{70}+u^{72} \right )du\\ &=\dfrac{-u^{47}}{47}+\dfrac{13u^{49}}{49}-\dfrac{78u^{51}}{51}+\dfrac{286u^{53}}{53}-\dfrac{715u^{55}}{55}+\dfrac{1287u^{57}}{57}-\frac{1716u^{59}}{59}+\dfrac{1716u^{61}}{61}-\dfrac{1287u^{63}}{63}+\dfrac{715u^{65}}{65}-\dfrac{286u^{67}}{67}+\dfrac{78u^{69}}{69}-\dfrac{13u^{71}}{71}+\dfrac{u^{73}}{73}\\ &=\dfrac{-u^{47}}{47}+\dfrac{13u^{49}}{49}-\dfrac{26u^{51}}{17}+\dfrac{286u^{53}}{53}-13u^{55}+\dfrac{429u^{57}}{19}-\dfrac{1716u^{59}}{59}+\dfrac{1716u^{61}}{61}-\dfrac{143u^{63}}{7}+11u^{65}-\dfrac{286u^{67}}{67}+\dfrac{26u^{69}}{23}-\dfrac{13u^{71}}{71}+\dfrac{u^{73}}{73}+C\\ \end{align} {/eq}

Reversing the substitution, we have:

{eq}\color{blue}{\int \sin ^{27}\left(x\right)\cos ^{46}\left(x\right)dx =\dfrac{-\cos^{47}(x)}{47}+\dfrac{13\cos^{49}(x)}{49}-\dfrac{26\cos^{51}(x)}{17}+\dfrac{286\cos^{53}(x)}{53}-13\cos^{55}(x)+\dfrac{429\cos^{57}(x)}{19}-\dfrac{1716\cos^{59}(x)}{59}+\dfrac{1716\cos^{61}(x)}{61}-\dfrac{143\cos^{63}(x)}{7}+11\cos^{65}(x)-\dfrac{286\cos^{67}(x)}{67}+\dfrac{26\cos^{69}(x)}{23}-\dfrac{13\cos^{71}(x)}{71}+\dfrac{\cos^{73}(x)}{73}+C} {/eq}

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13