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Question: \(\int_{}^{}\sqrt{\frac{1 - x}{1 + x}}\)· \(\frac{1}{x}\)dx is equal to –...

∫1−x1+x\int_{}^{}\sqrt{\frac{1 - x}{1 + x}}· 1x\frac{1}{x}dx is equal to –

A

–2 log ∣1+x+1−x1−x−1+x∣\left| \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 - x} - \sqrt{1 + x}} \right| + cos–1 x + c

B

– log ∣1−x+1+x1+x−1−x∣\left| \frac{\sqrt{1 - x} + \sqrt{1 + x}}{\sqrt{1 + x} - \sqrt{1 - x}} \right| + cos–1 x + c

C

– log ∣1+1−x21−1−x2∣\left| \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} \right|+ cos–1 x + c

D

None of these

Answer

– log ∣1+1−x21−1−x2∣\left| \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} \right|+ cos–1 x + c

Explanation

Solution

On putting x = cos 2q, and dx = –2 sin 2q dq,

We get I = ∫1−x1+x\int \sqrt { \frac { 1 - \mathrm { x } } { 1 + \mathrm { x } } } 1x\frac { 1 } { x } dx

= ∫1−cos⁡2θ1+cos⁡2θ\int \sqrt { \frac { 1 - \cos 2 \theta } { 1 + \cos 2 \theta } } × 1cos⁡2θ\frac { 1 } { \cos 2 \theta } × (–2 sin 2q) dq

= ∫2sin⁡2θ2cos⁡2θ\int \sqrt { \frac { 2 \sin ^ { 2 } \theta } { 2 \cos ^ { 2 } \theta } } × (−4sin⁡θcos⁡θcos⁡2θ)\left( \frac { - 4 \sin \theta \cos \theta } { \cos 2 \theta } \right) dq

= –∫4sin⁡2θcos⁡2θ\int \frac { 4 \sin ^ { 2 } \theta } { \cos 2 \theta }dq

= –2∫1−cos⁡2θcos⁡2θ\int \frac { 1 - \cos 2 \theta } { \cos 2 \theta }dq

= –2 (sec 2q – 1) dq

= –2 log |sec 2q + tan 2q| + 2q + c

= –2 log ∣1+sin⁡2θcos⁡2θ∣\left| \frac { 1 + \sin 2 \theta } { \cos 2 \theta } \right| + 2q + c

= –2 log ∣1+sin⁡2θ1−sin⁡22θ∣\left| \frac { 1 + \sin 2 \theta } { \sqrt { 1 - \sin ^ { 2 } 2 \theta } } \right| + 2q + c

= –2 log ∣1+sin⁡2θ1−sin⁡2θ∣\left| \sqrt { \frac { 1 + \sin 2 \theta } { 1 - \sin 2 \theta } } \right| + 2q + c

= – log ∣1+1−x21−1−x2∣\left| \frac { 1 + \sqrt { 1 - x ^ { 2 } } } { 1 - \sqrt { 1 - x ^ { 2 } } } \right| + cos–1 x + c

Hence (3) is the correct answer.