For x² ≠ nπ + 1, n N(the set of natural numbers) . the integral {2(x²-1)- 2(x²-1)/2(x²-1)+ 2(x²-1)} \, dx is equal to : (where c is a constant of integration) [9-Jan-2019 Shift 1]

Correct Answer is : log⁡e∣sec⁡(x2−12)∣_{e} ≤ft| ≤ft( x²-1/2 ) |logesec(2x2−1) + c

_{e} ≤ft| ≤ft( x²-1/2 ) | + c

1/2 _{e} ≤ft| ≤ft( x²-1/2 ) | + c

Correct Answer is : log⁡e∣sec⁡(x2−12)∣_{e} ≤ft| ≤ft( x²-1/2 ) |logesec(2x2−1) + c

Test Index

Indefinite Integrals

Section: Mathematics

Question : 72 of 90

Marks: +1, -0

For

x^2 \neq n\pi + 1, n \in N

\int \sqrt{\frac{2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}} \, dx

Solution:

From question, the integral given can be taken as I:

\Rightarrow I = \int \sqrt{\frac{2\sin(x^2-1)-\sin 2(x^2-1)}{2\sin(x^2-1)+\sin 2(x^2-1)}} \, dx

\because \frac{x^2-1}{2} = \theta

\Rightarrow x^2-1 = 2\theta

\Rightarrow 2x \frac{dx}{d\theta} = 2

\Rightarrow x \frac{dx}{d\theta} = 1

\therefore x \, dx = d\theta

\Rightarrow I = \int \sqrt{\frac{2\sin 2\theta - \sin 2(2\theta)}{2\sin 2\theta + \sin 2(2\theta)}} \, d\theta

\Rightarrow I = \int \sqrt{\frac{2\sin 2\theta - \sin 4\theta}{2\sin 2\theta + \sin 4\theta}} \, d\theta

\because [\sin 2\theta = 2\sin\theta\cos\theta \Rightarrow \sin 4\theta = \sin 2(2\theta) = 2\sin 2\theta \cos 2\theta]

\Rightarrow I = \int \sqrt{\frac{2\sin 2\theta - 2\sin 2\theta \cos 2\theta}{2\sin 2\theta + 2\sin 2\theta \cos 2\theta}} \, d\theta

\Rightarrow I = \int \sqrt{\frac{2\sin 2\theta(1-\cos 2\theta)}{2\sin 2\theta(1+\cos 2\theta)}} \, d\theta

\Rightarrow I = \int \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} \, d\theta

\because [1-\cos 2\theta = 2\sin^2\theta]

\because [1+\cos 2\theta = 2\cos^2\theta]

\Rightarrow I = \int \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} \, d\theta

\Rightarrow I = \int \sqrt{\tan^2\theta} \, d\theta

\Rightarrow I = \int (\tan\theta) \, d\theta

\Rightarrow I = \log_{e}|\sec\theta| + C

Now, putting the value of

\theta

\therefore I = \log_{e} \left| \sec\left( \frac{x^2-1}{2} \right) \right| + C

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