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Question -

$\int_0^{\frac{\pi}{2}} \frac{dx}{1+\cos^2x}$

I put $\cos x = t$

Then $-\sin x\; dx = dt$

But then I am totally confused what to do with $\sin x.$

How to solve this.

  • 0
    Try putting $t=\tan x$ – 2017-02-05
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    If you have mathematica, then a wolfram alpha query (with == in front of Integrate[1/(1 + Cos[x]^2), {x, 0, Pi/2}]) gives you a step by step solution. – 2017-02-05
  • 1
    I'm not sure the Fundamental Theorem applies to questions. – 2017-02-05

5 Answers 5

1

Well, when you substitute $t=\cos\left(x\right)$ then we know that:

$$\cos^2\left(x\right)+\sin^2\left(x\right)=t^2+\sin^2\left(x\right)=1\tag1$$

Another way, substitute $\text{u}=\tan\left(x\right)$:

$$\int_0^\frac{\pi}{2}\frac{1}{1+\cos^2\left(x\right)}\space\text{d}x=\int_0^\infty\frac{1}{2+\text{u}^2}\space\text{d}\text{u}\tag2$$

And after that substitute $\text{v}=\frac{\text{u}}{\sqrt{2}}$:

$$\int_0^\infty\frac{1}{2+\text{u}^2}\space\text{d}\text{u}=\frac{1}{\sqrt{2}}\int_0^\infty\frac{1}{1+\text{v}^2}\space\text{d}\text{v}=\frac{1}{\sqrt{2}}\left(\lim_{\text{v}\to\infty}\arctan\left(\text{v}\right)-\arctan\left(0\right)\right)=\frac{\frac{\pi}{2}-0}{\sqrt{2}}\tag3$$

  • 0
    How you fill limits and what should be the new limits? – 2017-02-05
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    @JohnSr See my edit! – 2017-02-05
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    What about $tan^{-1}\frac{\infty}{2}$? – 2017-02-05
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    @JohnSr You can not say that, because $\infty$ is not a number. You can take the limit: $$\lim_{x\to\infty}\arctan\left(\frac{x}{2}\right)=\frac{\pi}{2}$$ – 2017-02-05
1

We have $$I = \int_{0}^{\frac {\pi}{2}} \frac {1}{1+\cos^2 x} \mathrm{d}x = \int_{0}^{\frac {\pi}{2}} \frac {1}{\tan^2 x +2} \sec^2 x \mathrm {d}x = \int_{0}^{\infty} \frac {1}{u^2+2} \mathrm {d}u $$ by substituting $u = \tan x $. Hope you can take it from here. If you want to check, the answer is $\boxed {\frac {\pi}{2\sqrt {2}}} $.

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    How you fill limits? – 2017-02-05
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    @JohnSr $x$ tends from $0$ to $\pi/2$. We know $u =\tan x $. So $u $ tends from $\tan 0$ to $\tan \pi /2$, that is, from, $0$ to $\infty $. – 2017-02-05
1

Note $\displaystyle \cos x=\frac{1}{\sec x}$.

So, $$ \begin{align}I&=\int\frac{1}{\cos^2x+1}dx\\ &=\int\frac{\sec^2x}{\sec^2x+1}dx \\&=\int\frac{\sec^2x}{\tan^2x+2}dx \tag{1}\end{align}$$

Now Let $u=\tan x\rightarrow du=\sec^2xdx$ and substituting in $(1)$, we get

$$\begin{align}I&=\int\frac{du}{u^2+2}\\&=\frac{1}{\sqrt2}\cdot\arctan{\frac{u}{\sqrt2}}+C\\&=\frac{1}{\sqrt2}\cdot\arctan{\frac{\tan x}{\sqrt2}}+C\end{align}$$

Finally find the value by improper integral.

$$\begin{align}\int_0^\frac{\pi}{2}\frac{1}{\cos^2x+1}dx&=\lim_{t\rightarrow\frac{\pi}{2}^-}\frac{1}{\sqrt2}\cdot\arctan{\frac{\tan t}{\sqrt2}}-0\\&=\frac{1}{\sqrt2}\cdot\frac{\pi}{2}\\&=\boxed{\frac{\pi}{2\sqrt2}}\end{align}$$

  • 0
    How you fill limits and what should be the new limits? – 2017-02-05
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    I edited in my answer. see my edit! – 2017-02-05
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    What about $tan^{-1}\frac{\infty}{2}$? – 2017-02-05
  • 0
    Let $\displaystyle u=\frac{\tan t}{\sqrt2}$. Then $$ \lim_{t\rightarrow\frac{\pi}{\sqrt2}^-} \arctan\frac{\tan t}{\sqrt2}=\lim_{u\rightarrow\infty}\arctan u=\frac{\pi}{2}$$ – 2017-02-05
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    When $u$ tends to infinity, $\arctan u$ will tend to $\frac{\pi}{2}$. you can see arctan graph. – 2017-02-05
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    I can't edit t tends to $\frac{\pi}{2}^-$. – 2017-02-05
0

Bioche's rules prompt you to set $i=\tan x$, $\mathrm d t=(1+t^2)\mathrm d x$, so you finally get the integral of the rational function $$\int_0^\infty\frac{\mathrm dt}{2+t^2}.$$

-1

From $\cos x=t$ and $-\sin x\,dx=dt$ you draw

$$dx=-\frac{dt}{\sin x}=-\frac{dt}{\sqrt{1-t^2}}.$$

Then your integral becomes

$$-\int_1^{0}\frac{dt}{(1+t^2)\sqrt{1-t^2}}$$ which, unfortunately, doesn't look much simpler.

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    Then any other simple way? – 2017-02-05
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    @JohnSr: I am answering "what to do with $\sin x$". For the solution, the other answers are sufficient. – 2017-02-05
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    But i have confusion about limits specially what about $tan^{-1}\frac{\infty}{2}$? – 2017-02-05
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    @JohnSr: this doesn't appear in my post, don't ask here. – 2017-02-05