Solveeit Logo
Login

Question

Question: The integral \[\int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} \] is equ...

The integral ∫xcos⁡−1(1−x21+x2)dx\int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} is equal to:
Note(x>0)(x > 0)
A. −x+(1+x2)cot⁡−1x+C - x + (1 + {x^2}){\cot ^{ - 1}}x + C
B. −x+(1+x2)tan⁡−1x+C - x + (1 + {x^2}){\tan ^{ - 1}}x + C
C. −x−(1+x2)tan⁡−1x+C - x - (1 + {x^2}){\tan ^{ - 1}}x + C
D. x−(1+x2)cot⁡−1x+Cx - (1 + {x^2}){\cot ^{ - 1}}x + C

Explanation

Solution

According to given in the question we have to determine the integral ∫xcos⁡−1(1−x21+x2)dx\int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} . So, first of all we have to let x as tan⁡θ\tan \theta and then we have to find the differentiation of the expression we just let.

Formula used: ddxtan⁡x=sec⁡2x...................(A)\dfrac{d}{{dx}}\tan x = {\sec ^2}x...................(A)
cos⁡2θ=(1−x21+x2).....................(B)\cos 2\theta = \left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right).....................(B)
cos⁡−1(cos⁡θ)=θ.....................(C){\cos ^{ - 1}}(\cos \theta ) = \theta .....................(C)
Now, we have to let the given expression as I and then we have to substitute the all values we let and differentiated.
Now, as we know to integrate two functions or more than two functions we have to use the integration by part method hence, to solve the integral we have to use the integration by the help of by part method which is as below:
⇒∫uvdx=u∫vdx−∫(dudx∫vdx)dx.......................(A)\Rightarrow \int {uvdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)dx.......................(A)} } }
Now, with the help of the formula (A) above, we can easily solve the given integral using by parts method and we have to take 2θ2\theta as u and tan⁡θsec⁡2θ\tan \theta {\sec ^2}\theta as v according to L’ Hospital rule.
Now, we have to solve the obtained integration and then we have to substitute the value of θ\theta by x as we have let in the beginning of the solution.

Complete step-by-step solution:
Step 1: First of all we have to let x as tan⁡θ\tan \theta and then we have to find the differentiation of the expression we just let as mentioned in the solution hint. Hence,
⇒x=tan⁡θ\Rightarrow x = \tan \theta
Now, we have to differentiate the functions as obtained just above.
⇒ddθx=ddθtan⁡θ\Rightarrow \dfrac{d}{{d\theta }}x = \dfrac{d}{{d\theta }}\tan \theta
Now, to solve the differentiation we have to use the formula (A) as mentioned in the solution hint.
⇒x=sec⁡2θdθ..................(1)\Rightarrow x = {\sec ^2}\theta d\theta ..................(1)
Step 2: Now, we have to let the given integral as I mentioned in the solution hint. Hence,
I=∫xcos⁡−1(1−x21+x2)dx.......................(2)I = \int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} .......................(2)
Step 3: Now, we have to substitute the expression (1) as obtained in the solution step 1 into the expression (2) as obtained in the solution step 2. Hence,
I=∫tan⁡θcos⁡−1(1−x21+x2)sec⁡2θdθ.......................(3)I = \int {\tan \theta {{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right){{\sec }^2}\theta d\theta } .......................(3)
Step 4: Now, to simplify the expression (3) as obtained in the solution hint we have to use the formula (B) as mentioned in the solution hint. Hence,
I=∫tan⁡θcos⁡−1(cos⁡2θ)sec⁡2θdθ.......................(4)I = \int {\tan \theta {{\cos }^{ - 1}}\left( {\cos 2\theta } \right){{\sec }^2}\theta d\theta } .......................(4)
Step 5: Now, to solve the expression(4) as obtained in the solution step 4 we have to use the formula (C) as mentioned in the solution hint. Hence,
I=∫2θtan⁡θsec⁡2θdθ.......................(5)I = \int {2\theta \tan \theta {{\sec }^2}\theta d\theta } .......................(5)
Step 6: Now, to solve or find the integration of the expression (5) as obtained in the solution step 5 we have to use the formula (D) which is as mentioned in the solution hint. Hence,
As explained in the solution hint, we have to take 2θ2\theta as u and tan⁡θsec⁡2θ\tan \theta {\sec ^2}\theta as v according to L’ Hospital rule. Hence,

I=2θ∫tan⁡θsec⁡2θdθ−∫(d2θdθ∫tan⁡θsec⁡2θdθ)dθ I=2θtan⁡2θ2−2×12∫tan⁡2θdθ  I = 2\theta \int {\tan \theta {{\sec }^2}\theta d\theta } - \int {\left( {\dfrac{{d2\theta }}{{d\theta }}\int {\tan \theta {{\sec }^2}\theta d\theta } } \right)} d\theta \\\ I = 2\theta \dfrac{{{{\tan }^2}\theta }}{2} - 2 \times \dfrac{1}{2}\int {{{\tan }^2}\theta d\theta } \\\

Now, on eliminating the terms which can be eliminated,
⇒θtan⁡2θ−[∫(−1+sec⁡2θ)dθ]\Rightarrow \theta {\tan ^2}\theta - \left[ {\int {( - 1 + {{\sec }^2}\theta )d\theta } } \right]
Now, we have to find the integration if the terms left to integrate in the expression as obtained just above,
⇒θtan⁡2θ+θ−tan⁡θ+C ⇒θ(1+tan⁡2θ)−tan⁡θ+C..................(6)  \Rightarrow \theta {\tan ^2}\theta + \theta - \tan \theta + C \\\ \Rightarrow \theta (1 + {\tan ^2}\theta ) - \tan \theta + C..................(6) \\\
Step 7: Now, we have to substitute the value of θ\theta in the expression (6) as obtained in the solution step 6. Hence,
I=−x+(1+x2)tan⁡−1x+CI = - x + (1 + {x^2}){\tan ^{ - 1}}x + C
Hence, with the help of the formulas (A), (B), (C), and (D) we have determined the value of the given integral ∫xcos⁡−1(1−x21+x2)dx\int {x{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} =−x+(1+x2)tan⁡−1x+C = - x + (1 + {x^2}){\tan ^{ - 1}}x + C where, C is the constant.

Therefore option (B) is correct.

Note: To determine the integration of the given integral we have to let the term of the integral which is x as tan⁡θ\tan \theta and then we have to find the differentiation of the expression just obtained.
If there are two or more than two functions/terms to integrate then we have to use the L’ Hospital rule to integrate which is already explained in the solution hint.