How do you evaluate the integral $\int {cos}^{3} x \sqrt{sin x}$ $int cos^3xsqrtsinx$ ?

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How do you evaluate the integral $\int {cos}^{3} x \sqrt{sin x}$ $int cos^3xsqrtsinx$ ?

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Answer 1 · 2017-01-21T17:03:03

$\frac{2}{3} {(sin x)}^{\frac{3}{2}} - \frac{2}{7} {(sin x)}^{\frac{7}{2}} + C$ $2/3(sinx)^(3/2) - 2/7(sinx)^(7/2) + C$

Explanation:

Let $u = sin x$ $u = sinx$ .

Then $d u = cos x d x$ $du = cosxdx$ and $d x = \frac{d u}{cos x}$ $dx = (du)/cosx$

$\Rightarrow \int {cos}^{3} x \sqrt{u} \cdot \frac{d u}{cos x}$ $=>intcos^3xsqrt(u) * (du)/cosx$

$\Rightarrow \int {cos}^{2} x \sqrt{u} d u$ $=>int cos^2x sqrt(u)du$

We can rewrite ${cos}^{2} x$ $cos^2x$ as $1 - {sin}^{2} x$ $1 - sin^2x$ .

$\Rightarrow \int (1 - {sin}^{2} x) \sqrt{u} d u$ $=>int (1 - sin^2x)sqrt(u)du$

Since $u = sin x$ $u = sinx$ , $u^{2} = {sin}^{2} x$ $u^2 = sin^2x$

$\Rightarrow \int (1 - u^{2}) \sqrt{u} d u$ $=>int(1 - u^2)sqrt(u)du$

$\Rightarrow \int (1 - u^{2}) u^{\frac{1}{2}} d u$ $=> int(1 - u^2)u^(1/2)du$

$\Rightarrow \int u^{\frac{1}{2}} - u^{\frac{5}{2}} d u$ $=>int u^(1/2) - u^(5/2)du$

This can be evaluate as $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n + 1)/(n + 1) + C$

$\Rightarrow \frac{2}{3} u^{\frac{3}{2}} - \frac{2}{7} u^{\frac{7}{2}} + C$ $=>2/3u^(3/2) - 2/7u^(7/2) + C$

$\Rightarrow \frac{2}{3} {(sin x)}^{\frac{3}{2}} - \frac{2}{7} {(sin x)}^{\frac{7}{2}} + C$ $=>2/3(sinx)^(3/2) - 2/7(sinx)^(7/2) + C$

Hopefully this helps!

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How do you evaluate the integral $\int {cos}^{3} x \sqrt{sin x}$ $int cos^3xsqrtsinx$ ?

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