My Math Forum plz help me to solve this problem

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 May 15th, 2017, 02:43 AM #1 Newbie   Joined: May 2017 From: qatar Posts: 6 Thanks: 0 plz help me to solve this problem
 May 15th, 2017, 03:12 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,191 Thanks: 871 Remembering the trig identity $1+ tan^2(\theta)= sec^2(\theta)$, I would let $x= tan(\theta)$. Then $dx= sec^2(\theta) d\theta$ so the integral becomes $\int sec(\theta)(sec^2(\theta)d\theta)= \int sec^3(\theta)d\theta$. Now you should remember that $sec(\theta)$ is defined as $\frac{1}{cos(\theta)$ so that integral is the same as $\int\frac{d\theta}{cos^3(\theta)}$. That has cosine to an odd power and there is a standard way of treating such integrals- multiply both numerator and denominator by $cos(\theta)$ to get $\int \frac{cos(\theta) d\theta}{cos^4(\theta)}= \int \frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}$. Let $u= sin(\theta)$ so that $du= cos(\theta)d\theta$ and the integral becomes $\int\frac{du}{(1- u^2)^2}= \int\frac{du}{(1- u)^2(1+ u)^2}$ which can be done by "partial fractions". Last edited by Country Boy; May 15th, 2017 at 03:15 AM.
May 15th, 2017, 03:51 AM   #3
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Quote:
 Originally Posted by senawy Attachment 8842
I do not believe this function has an elementary, closed-form antiderivative.

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May 15th, 2017, 05:07 AM   #4
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Quote:
 Originally Posted by Country Boy Remembering the trig identity $1+ tan^2(\theta)= sec^2(\theta)$, I would let $x= tan(\theta)$. Then $dx= sec^2(\theta) d\theta$ so the integral becomes $\int sec(\theta)(sec^2(\theta)d\theta)= \int sec^3(\theta)d\theta$. Now you should remember that $sec(\theta)$ is defined as $\frac{1}{cos(\theta)$ so that integral is the same as $\int\frac{d\theta}{cos^3(\theta)}$. That has cosine to an odd power and there is a standard way of treating such integrals- multiply both numerator and denominator by $cos(\theta)$ to get $\int \frac{cos(\theta) d\theta}{cos^4(\theta)}= \int \frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}$. Let $u= sin(\theta)$ so that $du= cos(\theta)d\theta$ and the integral becomes $\int\frac{du}{(1- u^2)^2}= \int\frac{du}{(1- u)^2(1+ u)^2}$ which can be done by "partial fractions".
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