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$$\int 5 (\tan(\sin(x)))^3 \cos(x)dx$$

What I did was let $\sin x=u$, then it's

$$\int 5\tan(u)^3 du$$

then I broke it up into parts

$$5\int(\tan(u))^2 \tan(u) du$$

$$5\int((\sec(u))^2-1)) \tan(u) du$$

$$5\int(\sec(u))^2(\tan(u))-\tan(u))du$$

then I let $(\sec(u))^2 = v$

$$5\int(v)dv+\ln|\cos(u)|$$

replaced the variables in terms of $x$

$$\frac{5}{2}(\sec(\sin(x)))^4 + \ln|\cos(\sin(x))|$$

This is incorrect but I do not see why. Thank you.

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    $\begingroup$ Type in Latex please! $\endgroup$ – Extremal Feb 19 '15 at 23:04
  • $\begingroup$ I don't know it!! I am sorry! $\endgroup$ – King Squirrel Feb 19 '15 at 23:04
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    $\begingroup$ For your second substitution, you should have let $v=\tan u$. $\endgroup$ – user170231 Feb 19 '15 at 23:06
  • $\begingroup$ It has been edited. $\endgroup$ – King Squirrel Feb 19 '15 at 23:08
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We have \begin{align} \int 5\tan^3\left(\sin\left(\theta\right)\right)\cos\left(\theta\right)\:d\theta,\tag{1} \end{align} and by letting $u=\sin\left(\theta\right),\:du=\cos\left(\theta\right)\:d\theta\implies\displaystyle\frac{du}{\cos\left(\theta\right)}=d\theta$, therefore giving us \begin{align} 5\int \tan^3\left(u\right)\cos\left(\theta\right)\frac{du}{\cos\left(\theta\right)}&=5\int\tan^3\left(u\right)\:du\tag{2}, \end{align} and now by the common reduction of powers s. th. \begin{align} \int \tan^n\left(\theta\right)\:d\theta=\frac{\tan^{n-1}\left(\theta\right)}{n-1}-\int\tan^{n-2}\left(\theta\right)\:d\theta, \end{align} we find that \begin{align} 5\int\tan^3\left(u\right)\:du&=5\left\{\frac{\tan^2\left(u\right)}{2}-\int\tan\left(u\right)\:du\right\}\tag{3}\\ &=\frac{5\tan^2\left(u\right)}{2}-5\int\frac{\sin\left(u\right)}{\cos\left(u\right)}\:du\tag{4}\\ &=\frac{5\tan^2\left(u\right)}{2}+5\log\left(\cos\left(u\right)\right)+C_1\tag{5}\\ &=\frac{5\tan^2\left(\sin\left(\theta\right)\right)}{2}+5\log\left(\cos\left(\sin\left(\theta\right)\right)\right)+C_1\tag{6}. \end{align} Alternatively, back at $\left(2\right)$ we could instead go to \begin{align} 5\int\tan^3\left(u\right)\:du&=5\int \tan\left(u\right)\left[\sec^2\left(u\right)-1\right]\:du\tag{7}\\ &=5\int\tan\left(u\right)\sec^2\left(u\right)\:du-5\int\tan\left(u\right)\:du\tag{8}\\ &=\frac{5\tan^2\left(u\right)}{2}+5\log\left(\cos\left(u\right)\right)+C_2,\tag{9} \end{align} and then by substituting back in for $u$ to get \begin{align} \int 5\tan^3\left(\sin\left(\theta\right)\right)\cos\left(\theta\right)\:d\theta&=\frac{5\tan^2\left(\sin\left(\theta\right)\right)}{2}+5\log\left(\cos\left(\sin\left(\theta\right)\right)\right)+C_2,\tag{10} \end{align} which is the same as $\left(6\right)$.

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  • $\begingroup$ Thank you great sir, I have it now and understand it completely. $\endgroup$ – King Squirrel Feb 19 '15 at 23:38
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Indeed that is incorrect because of the substitution $v=(sec\ u)^2$. If $v$ is such then $dv$ must be? $2sec\ u\cdot sec\ u\ tan\ u=2(sec\ u)^2 tan\ u$.

Looking on the integral above: $5\int (v)dv=5\int (sec\ u)^2\cdot 2(sec\ u)^2 tan\ u\not =5\int (sec\ u)^2tan\ u$

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You could proceed as follows (I left out the factor 5): $$\int \tan^3 u \, du = \int \tan u (\sec^2u-1)\, du = \int \tan u\sec^2u \, du-\int \tan u\, du$$ The last one is easy ($\tan x = \sin x / \cos x$, let $t = \cos x$) and for the first one, choose $t = \tan u$, then $dt = \sec² u \, du$.

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Hint: $~\displaystyle\int\tan^3x~dx=\int\frac{\sin^3x}{\cos^3x}dx=-\int\frac{1-\cos^2x}{\cos^3x}~d(\cos x).$

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