Question

How do you integrate $\sqrt{1-{\tan }^{2}x}$ with respect to $$x$$ ?

Answer 1

Hint: In this question we have to find the integral of the given function, first multiply and divide with $${\sec }^{2}x$$ then assume the $$\tan x$$ as $$\sin \theta$$ , and then simplify the expression by using partial fractions, then we will get an expression that is easy to be integrated, so apply integration formula we will get the required value for the given expression.

Complete step by step answer:
Given $$\sqrt{1-{\tan }^{2}x}$$ and we have to integrate with respect to $$x$$ ,
Let,
 $$I=\int \sqrt{1-{\tan }^{2}x}$$ ,
Multiply and divide with $${\sec }^{2}x$$ , we get,
$$\Rightarrow I=\int \sqrt{1-{\tan }^{2}x}\times {\displaystyle \frac{{\sec }^{2}x}{{\sec }^{2}x}}dx$$ ,
Now we know that, $${\sec }^{2}x=1+{\tan }^{2}x$$ now substitute the value in the expression we get,
$$\Rightarrow I=\int {\displaystyle \frac{\sqrt{1-{\tan }^{2}x}}{1+{\tan }^{2}x}}\left({\sec }^{2}x\right)dx$$ ,
Let us consider $$\sin \theta =\tan x$$ ,
Now perform differentiate on both sides, we get,
$$\Rightarrow {\displaystyle \frac{d}{d\theta }}\sin \theta ={\displaystyle \frac{d}{dx}}\tan x$$ ,
Now deriving we get,
$$\Rightarrow \cos \theta d\theta ={\sec }^{2}xdx$$ ,
Now the given expression becomes,
$$\Rightarrow I=\int {\displaystyle \frac{\sqrt{1-{\sin }^2\theta }}{1+{\sin }^2\theta }}\cos \theta d\theta$$ ,
We know that $$1-{\sin }^2\theta ={\cos }^2\theta$$ , then the expression becomes,
$$\Rightarrow I=\int {\displaystyle \frac{\sqrt{{\cos }^2\theta }}{1+{\sin }^2\theta }}\cos \theta d\theta$$ ,
Now simplifying we get,
$$\Rightarrow I=\int {\displaystyle \frac{\cos \theta }{1+{\sin }^2\theta }}\cos \theta d\theta$$ ,
Now multiplying we get,
$$\Rightarrow I=\int {\displaystyle \frac{{\cos }^2\theta }{1+{\sin }^2\theta }}d\theta$$ ,
Now writing in terms of $$\tan \theta$$ and $$\sec \theta$$ , we get,
$$\Rightarrow I=\int {\displaystyle \frac{{\displaystyle \frac{1}{{\sec }^2\theta }}}{1+{\displaystyle \frac{{\tan }^2\theta }{{\sec }^2\theta }}}}d\theta$$ ,
Now simplifying we get,
$$\Rightarrow I=\int {\displaystyle \frac{1}{{\sec }^2\theta +{\tan }^2\theta }}d\theta$$ ,
Now we know that $$1+{\tan }^2\theta ={\sec }^2\theta$$ , we get,
$$\Rightarrow I=\int {\displaystyle \frac{1}{1+{\tan }^2\theta +{\tan }^2\theta }}d\theta$$ ,
Now simplifying we get,
$$\Rightarrow I=\int {\displaystyle \frac{1}{1+2{\tan }^2\theta }}d\theta$$ ,
Now multiply and divide by $${\sec }^2\theta$$ , we get,
$$\Rightarrow I=\int {\displaystyle \frac{1}{1+2{\tan }^2\theta }}\times {\displaystyle \frac{{\sec }^2\theta }{{\sec }^2\theta }}d\theta$$ ,
Now using the identity $$1+{\tan }^2\theta ={\sec }^2\theta$$ , we get,
$$\Rightarrow I=\int {\displaystyle \frac{1}{1+2{\tan }^2\theta }}\times {\displaystyle \frac{{\sec }^2\theta }{1+{\tan }^2\theta }}d\theta$$ ,
Now simplifying we get,
$$\Rightarrow I=\int {\displaystyle \frac{{\sec }^2\theta }{\left(1+{\tan }^2\theta \right)\left(1+2{\tan }^2\theta \right)}}d\theta$$ ,
Let us consider $$u=\tan \theta$$ ,
Now deriving on both sides we get,
$$\Rightarrow du={\sec }^2\theta d\theta$$ ,
Now the expression becomes,
$$\Rightarrow I=\int {\displaystyle \frac{1}{\left(1+u^2\right)\left(1+2u^2\right)}}du$$ ,
Now using performing partial fractions, we get, rewrite the expression we get,
$$\Rightarrow {\displaystyle \frac{A}{1+u^2}}+{\displaystyle \frac{B}{1+2u^2}}={\displaystyle \frac{1}{\left(1+u^2\right)\left(1+2u^2\right)}}$$ ,
Now taking L.C.M and simplifying we get,
$$\Rightarrow {\displaystyle \frac{A+2A{u}^2+B+B{u}^2}{\left(1+u^2\right)\left(1+2u^2\right)}}={\displaystyle \frac{1}{\left(1+u^2\right)\left(1+2u^2\right)}}$$ ,
Now simplifying we get,
$$\Rightarrow {\displaystyle \frac{u^2\left(2A+B\right)+\left(A+B\right)}{\left(1+u^2\right)\left(1+2u^2\right)}}={\displaystyle \frac{1}{\left(1+u^2\right)\left(1+2u^2\right)}}$$ ,
Now comparing the two sides we get,
 $$2A+B=0----(1)$$ and
 $$A+B=1----(2)$$ ,
Now solving the two equations we get,
$$\Rightarrow A=-1$$ and $$B=2$$ ,
Now substitute the values in the expression we get,
$$\Rightarrow I=\int {\displaystyle \frac{2}{1+2u^2}}-{\displaystyle \frac{1}{1+u^2}}du$$ ,
Now separating the terms we get,
$$\Rightarrow I=\int {\displaystyle \frac{2}{1+2u^2}}du-\int {\displaystyle \frac{1}{1+u^2}}du$$ ,
Now we can say that the above terms are in form of $${\tan }^{-1}$$ integrals,
$$\Rightarrow I=\sqrt{2}\int {\displaystyle \frac{\sqrt{2}}{1+\sqrt{2}{u}^2}}du-\int {\displaystyle \frac{1}{1+u^2}}du$$ ,
So, from the integral formula we get,
$$\Rightarrow I=\sqrt{2}{\tan }^{-1}\left(\sqrt{2}u\right)-{\tan }^{-1}u$$ ,
Now we know that $$u=\tan \theta$$ , substitute the value in the expression we get,
$$\Rightarrow I=\sqrt{2}{\tan }^{-1}\left(\sqrt{2}\tan \theta \right)-{\tan }^{-1}\left(\tan \theta \right)$$ ,
Now simplifying we get,
$$\Rightarrow I=\sqrt{2}{\tan }^{-1}\left(\sqrt{2}\tan \theta \right)-\theta$$ ,
Now we know that $$\sin \theta =\tan x$$ ,
Now apply $${\sin }^{-1}$$ on both sides we get,
$$\Rightarrow {\sin }^{-1}\left(\sin \theta \right)={\sin }^{-1}\left(\tan x\right)$$ ,
Now simplifying we get,
$$\Rightarrow \theta ={\sin }^{-1}\left(\tan x\right)$$ ,
And we know that $$\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$$ , and $$\cos \theta =\sqrt{1-{\sin }^2\theta }$$ , so we get,
$$\Rightarrow$$ $$\tan \theta ={\displaystyle \frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}}$$ ,
Now we know that $$\sin \theta =\tan x$$ , by substituting we get,
$$\Rightarrow \tan \theta ={\displaystyle \frac{\tan x}{\sqrt{1-{\tan }^{2}x}}}$$ ,
Now the expression becomes,
$$\Rightarrow I=\sqrt{2}{\tan }^{-1}\left({\displaystyle \frac{\sqrt{2}\tan x}{\sqrt{1-{\tan }^{2}x}}}\right)-{\sin }^{-1}\left(\tan x\right)+C$$ ,
Now substituting the value of I we get,
$$\Rightarrow \int \sqrt{1-{\tan }^{2}x}dx=\sqrt{2}{\tan }^{-1}\left({\displaystyle \frac{\sqrt{2}\tan x}{\sqrt{1-{\tan }^{2}x}}}\right)-{\sin }^{-1}\left(\tan x\right)+C$$ ,
So, the value of the given expression is,
 $$\sqrt{2}{\tan }^{-1}\left({\displaystyle \frac{\sqrt{2}\tan x}{\sqrt{1-{\tan }^{2}x}}}\right)-{\sin }^{-1}\left(\tan x\right)+C$$ ,

Final Answer:
 $$\therefore$$ The value of the given expression $$\int \sqrt{1-{\tan }^{2}x}dx$$ is equal to $$\sqrt{2}{\tan }^{-1}\left({\displaystyle \frac{\sqrt{2}\tan x}{\sqrt{1-{\tan }^{2}x}}}\right)-{\sin }^{-1}\left(\tan x\right)+C$$ .

Note: In this type of question we use both derivation and integration formulas as both are related to each other students should not get confused where to use the formulas and which one to use as there are many formulas in both derivation and integration. Some of the important formulas that are used are given below:
 $${\displaystyle \frac{d}{dx}}x=1$$ ,
 $${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}$$ ,
 $${\displaystyle \frac{d}{dx}}uv=u{\displaystyle \frac{dv}{dx}}+v{\displaystyle \frac{du}{dx}}$$ ,
 $${\displaystyle \frac{d}{dx}}{e}^x=e^x$$ ,
 $${\displaystyle \frac{d}{dx}}{\displaystyle \frac{u}{v}}={\displaystyle \frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}}$$ ,
 $${\displaystyle \frac{d}{dx}}\ln x={\displaystyle \frac{1}{x}}$$ ,
 $${\displaystyle \frac{d}{dx}}\sin x=\cos x$$ ,
 $${\displaystyle \frac{d}{dx}}\cos x=-\sin x$$ ,
 $${\displaystyle \frac{d}{dx}}\tan x={\sec }^{2}x$$ ,
 $$\int dx=x+c$$ ,
 $$\int x^{n}dx={\displaystyle \frac{x^{n+1}}{n+1}}+c$$ ,
 $$\int {\displaystyle \frac{1}{x}}dx=\ln x+c$$ ,
 $$\int {\displaystyle \frac{1}{1+x^2}}dx={\tan }^{-1}x$$ .