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How would I integrate to evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution?

I made an attempt by making substitutions such as $$\cos^2(x)=1-\sin^2(x)$$ $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$ and $$\tan^2(x)=1+\sec^2(x)$$ But I couldn't find a way to make it look like an integral I could solve using a $u$ substitution or identity. Could I get some help on this one?

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$$\int \cos^2x\tan^3x\ dx=\int \frac{\sin^3x}{\cos x}\ dx=\int \frac{(1-\cos^2x)\sin x}{\cos x}\ dx$$

Let $\cos x=t\implies -\sin x\ dx=dt$ $$=\int \frac{(t^2-1)dt}{t}$$ $$=\int \left(t-\frac{1}{t}\right)dt$$

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Use the definition of $\tan x$, then after a bit of algebra you have $$ \frac{\sin x }{\cos x} - \cos x \sin x $$ The first one is solved using logarithm subsitution, the second using the identity $2 \sin x \cos x = \sin 2 x$

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    $\begingroup$ For the integral sin cos, no need to use the identity you mentioned we can just use the power rule :) $\endgroup$ – Ali Shadhar Jun 1 at 21:36
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$$\int \left (\cos^{2}x \right )\cdot \left (\tan^{3}x \right )dx$$$$ = \int \left (\cos^{2}x \right )\cdot \left (\frac{\sin^{3}x }{\cos^{3}x} \right)dx$$ $$=\int \left (\frac{\sin^{3}x }{\cos x} \right )dx$$$$ = \int \left (\frac{\left (\sin^{2}x \right )\sin x}{cos x} \right )dx$$ $$= \int \left (\frac{\left (1- \cos^{2}x \right )\sin x}{\cos x} \right )dx$$$$= \int \left [\left (\frac{\sin x}{\cos x} \right )-\left ( \cos x\cdot \sin x \right ) \right ]dx= ...$$

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  • $\begingroup$ Is your answer $-\ln|\cos x|-\sin^2x/2$? $\endgroup$ – user766881 Jun 13 at 7:10
  • $\begingroup$ Yes, your answer and Maria's answer are the same up to a constant. $\endgroup$ – Toby Mak Jun 13 at 7:14
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Simplify: $$\cos^2x\tan^3x=\cos^2x\frac{\sin^3x}{\cos^3x}=\frac{\sin^3x}{\cos x}$$

Assume that $$\cos x=u, \ \ \ dx=-\frac{du}{\sin x}$$ $$\int \frac{\sin^3xdx}{\cos x}=\int \frac{\sin^3xdx}{u}\frac{-du}{\sin x}$$ $$=-\int \frac{(1-\cos^2x)}{u} du$$ $$=-\int \frac{(1-u^2)}{u} du$$ $$=\int (u-\frac{1}{u}) du$$ $$=\frac{u^2}{2}-\ln|u|+C$$ $$=\frac{\cos^2x}{2}-\ln|\cos x|+C$$ Where, $C$ is a constant for integration.

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Just for fun:

$$\int \cos^2 x \tan^3x \ \mathrm{d} x$$ $$=\int \frac{\tan^3 x}{\sec^2 x} \ \mathrm{d} x$$

Now let $u = \tan^2 x, \mathrm du = 2 \tan x \sec^2 x \ \mathrm dx$:

$$=\int \frac{u \tan x }{\sec^2 x} \cdot \frac{\mathrm{d} u}{2 \tan x \sec^2 x}$$ $$=\frac{1}{2} \int \frac{u}{\sec^4 x} \mathrm d u$$ $$=\frac{1}{2} \int \frac{u}{(1+u)^2} \ \mathrm d u$$ $$=\frac{1}{2} \int \frac{1+u}{(1+u)^2} -\frac{1}{(1+u)^2} \mathrm d u$$ $$=\frac{1}{2} \left( \ln(\tan^2 x + 1) + \frac{1}{1+\tan^2 x} \right) +C$$ $$=\frac{1}{2} \left( \ln | \sec^2 x| + \cos^2 x \right) + C$$

where we have used the identity $1 + \tan^2 x = \sec^2 x$ twice.

Further simplifying gives the accepted answer: $$=\frac{1}{2} \left( \ln | \cos^{-2} x| + \cos^2 x \right) + C$$ $$=\frac{1}{2} \left(-2 \ln | \cos x | + \cos^2 x \right) + C$$ $$=\frac{\cos^2 x}{2} - \ln | \cos x | + C$$

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    $\begingroup$ OK, that's good. Your answer is same as mine $\endgroup$ – user766881 Jun 13 at 7:22

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