Question #91a61

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Question #91a61

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Answer 1 · 2016-05-07T10:09:14

$- (csc x + sin x)$ $-(cscx+sinx)$

Explanation:

$\int {cos}^{3} x \cdot {csc}^{2} x d x = \int {cos}^{2} x \cdot cos x \cdot \frac{1}{{sin}^{2} x} d x$ $intcos^3x * csc^2x dx =intcos^2x*cosx *1/sin^2x dx$
$= \int (1 - {sin}^{2} x) \cdot \frac{cos x}{{sin}^{2} x} d x$ $=int(1-sin^2x)*cosx /sin^2x dx$

Now let $z = sin x \Rightarrow d z = cos x d x$ $z=sinx =>dz=cosxdx$

so changing the varable from x to z the integral becomes

$= \int \frac{1 - z^{2}}{z^{2}} d z = \int \frac{d z}{z^{2}} - \int d z = \frac{z^{- 2 + 1}}{- 2 + 1} - z + c$ $=int(1-z^2)/z^2 dz=intdz/z^2-intdz=z^(-2+1)/(-2+1)-z+c$
$= - \frac{1}{z} - z + c = - \frac{1}{sin x} - sin x = - (csc x + sin x)$ $=-1/z-z+c=-1/sinx-sinx=-(cscx+sinx)$

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Question #91a61

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