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Use the substitution $x = \tan\theta$ to show that $$\int\frac {1 - x^2} {(1+x^2)^2} dx = \int \cos2\theta\ d\theta $$

I'm a bit lost on how to handle this question, I have tried subtituting $d\theta/dx = 1 / \sec^2\theta$ but I still don't reach the answer.

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Alternatively, I recommend the simpler approach using algebra:

$$\int\frac{1-x^2}{1+x^2}\ dx=-\int\frac{x^2+1-2}{x^2+1}\ dx=-\int1\ dx+\int\frac2{x^2+1}\ dx$$

The first integral is trivial, and the second is clearly arctan, so

$$\int\frac{1-x^2}{1+x^2}\ dx=-x+2\arctan(x)+c$$

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    $\begingroup$ @the_candyman Algebra before any integral technique I always say... :-) $\endgroup$ – Simply Beautiful Art Mar 29 '17 at 18:56
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I think your identity is not true. with $$x=\tan(\theta)$$ we have $$dx=\frac{1}{\cos(\theta)^2}d\theta$$ we get for the integrand: $$\int\frac{ \frac{\cos(\theta)^2-\sin(\theta)^2}{\cos(\theta)^2}}{\frac{\sin(\theta)^2+\cos(\theta)^2}{\cos(\theta)^2}}\cdot \frac{1}{\cos(\theta)^2}d \theta=\int \frac{\cos(\theta)^2-\sin(\theta)^2}{\cos(\theta)^2}d \theta=\int\frac{\cos(2\theta)}{\cos^2(\theta)}d\theta$$

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  • $\begingroup$ Indeed, $\int\cos(2\theta)\ dx$ should be correct. $\endgroup$ – Simply Beautiful Art Mar 29 '17 at 19:03
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Try using trig identities $$\sin^2(x) = \frac{\tan^2(x)}{1+\tan^2(x)}$$ $$\cos^2(x) = \frac1{1+\tan^2(x)}$$ $$\cos^2(x)-\sin^2(x) = \cos(2x)$$

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I have done with your substitution yet it is so boring $$\int \frac { 1-x^{ 2 } }{ (1+x^{ 2 }) } dx=\int { \frac { 1-\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } \frac { 1 }{ \cos ^{ 2 }{ \theta } } d\theta =\int { \left( 1-\tan ^{ 2 }{ \theta } \right) d\theta =\int { \frac { \cos ^{ 2 }{ \theta -\sin ^{ 2 }{ \theta } } }{ \cos ^{ 2 }{ \theta } } } d\theta =\int { \frac { \cos { 2\theta } }{ \cos ^{ 2 }{ \theta } } d\theta } = } } \\ =\int { \cos { 2\theta \quad d\left( \tan { \theta } \right) } } $$ the last part we can evaluate by parts

$$\\ \\ \int { \cos { 2\theta \quad d\left( \tan { \theta } \right) } } =\cos { 2\theta \tan { \theta +2\int { \sin { 2\theta } \cdot \tan { \theta = } } } } \cos { 2\theta \tan { \theta +4\int { \sin ^{ 2 }{ \theta d\theta } } } } =\cos { 2\theta \tan { \theta +2\int { \left( 1-\cos { 2\theta } \right) d\theta } = } } \\ =\cos { 2\theta \tan { \theta +2\left( \theta -\frac { \sin { 2\theta } }{ 2 } \right) } } =\cos { 2\theta \tan { \theta +2\theta -\sin { 2\theta +C } } } $$

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