Integration Question (1 Viewer)

Prashant Sallan · Mar 29, 2009

integrate (sin cubed x / cos squared x) :burn:

lolokay · Mar 29, 2009

sin³x = sinx(1-cos²x) ..

Drongoski · Mar 29, 2009

Is the answer: cos x + sec x + C ??

Drongoski · Mar 29, 2009

Doing it your conventional way:

$Let\ \ u\ =\ cos\ x\ \ \ \therefore \ \ du\ =\ -sin\ x\ dx\\ \\ \int \frac{sin^{3}x}{cos^{2}x}\ dx\ \ \ =\ \ \int tan^{2}x\ .\ sin\ x\ dx\ \ =\ \ \int (sec^{2}x\ -\ 1)\ sin\ x\ dx\\ \\ \ \ =\ \ \int [(cos\ x)^{-2}\ -\ 1]\ sin\ x\ dx\ \ =\ \ \int (1\ -\ u^{-2})\ du\\ \\ \ \ =\ \ u\ +\ \frac{1}{u}\ +\ C\ \ =\ \ cos\ x\ \ +\ \ sec\ x\ \ +\ \ C$

Trebla · Mar 29, 2009

No need for substitution:
$\displaystyle\int\dfrac{\sin^3x}{\cos^2x}\,dx = \displaystyle\int\dfrac{\sin x(1-\cos^2x)}{\cos^2x}\,dx\\= \displaystyle\int \left (\dfrac{\sin x}{\cos^2x}-\sin x \right )\,dx\\= \displaystyle\int (\sec x\tan x -\sin x)\,dx\\= \sec x + \cos x + c$

Integration Question (1 Viewer)

Prashant Sallan

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lolokay

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Drongoski

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Drongoski

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Trebla

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