There exist θ such that a > ≤ft| θ |, then \; dx/1 + a x =

Correct Answer is : 1a2−1log⁡(a+1cos⁡x2+a−1sin⁡x2a+1cos⁡x2−a−1sin⁡x2)+C\; {1}{{a² - 1}} ≤ft( {{a + 1} x/2 + {a - 1} x/2}{{a + 1} x/2 - {a - 1} x/2} ) + Ca2−11log(a+1cos2x−a−1sin2xa+1cos2x+a−1sin2x)+C

\; {1}{{a² - 1}} ^{-1}≤ft( {{a - 1}}{{a + 1}} x/2 ) + C

\; {1}{{a² - 1}} ^{-1}≤ft( {1 - a/1 + a} x/2 ) + C

\; {1}{{a² - 1}} ≤ft( {{a + 1} x/2 - {a - 1} x/2}{{a - 1} x/2 + {a - 1} x/2} ) + C

\; {1}{{a² - 1}} ≤ft( {{a + 1} x/2 + {a - 1} x/2}{{a + 1} x/2 - {a - 1} x/2} ) + C

There exist θ such that a > ≤ft| θ |, then \; dx/1 + a x =

Correct Answer is : 1a2−1log⁡(a+1cos⁡x2+a−1sin⁡x2a+1cos⁡x2−a−1sin⁡x2)+C\; {1}{{a² - 1}} ≤ft( {{a + 1} x/2 + {a - 1} x/2}{{a + 1} x/2 - {a - 1} x/2} ) + Ca2−11log(a+1cos2x−a−1sin2xa+1cos2x+a−1sin2x)+C

Test Index

AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 70 of 160

Marks: +1, -0

There exist

\theta

such that

a > \left| \sec \theta \right|

, then

\int\; \frac{dx}{1 + a \cos x} =

$\; \frac{1}{\sqrt{a^2 - 1}} \tan^{-1}\left( \frac{\sqrt{a - 1}}{\sqrt{a + 1}} \tan \frac{x}{2} \right) + C$
$\; \frac{1}{\sqrt{a^2 - 1}} \tan^{-1}\left( \sqrt{\frac{1 - a}{1 + a}} \tan \frac{x}{2} \right) + C$
$\; \frac{1}{\sqrt{a^2 - 1}} \log\left( \frac{\sqrt{a + 1} \cos \frac{x}{2} - \sqrt{a - 1} \sin \frac{x}{2}}{\sqrt{a - 1} \cos \frac{x}{2} + \sqrt{a - 1} \sin \frac{x}{2}} \right) + C$
$\; \frac{1}{\sqrt{a^2 - 1}} \log\left( \frac{\sqrt{a + 1} \cos \frac{x}{2} + \sqrt{a - 1} \sin \frac{x}{2}}{\sqrt{a + 1} \cos \frac{x}{2} - \sqrt{a - 1} \sin \frac{x}{2}} \right) + C$

Validate

Solution: 👈: Video Solution

Let

I = \int\; \frac{dx}{1 + a \cos x}

= \int\; \frac{1}{1 + a \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)} \; dx

\; = \int\; \frac{\sec^2 \frac{x}{2} \, dx}{1 + \tan^2 \frac{x}{2} + a \left(1 - \tan^2 \frac{x}{2}\right)}

\; = \int\; \frac{\sec^2 \frac{x}{2} \, dx}{(1 + a) - \tan^2 \frac{x}{2} (a - 1)}

\; = \frac{1}{a + 1} \int\; \frac{\sec^2 \frac{x}{2}}{1 - \left( \frac{a - 1}{a + 1} \right) \tan^2 \frac{x}{2}} \, dx

\; \text{ Let } \sqrt{\frac{a - 1}{a + 1}} \tan \frac{x}{2} = t

\; \Rightarrow \sqrt{\frac{a - 1}{a + 1}} \times \frac{1}{2} \times \sec^2 \frac{x}{2} \, dx = dt

\; \Rightarrow \sec^2 \frac{x}{2} \, dx = 2 \sqrt{\frac{a + 1}{a - 1}} \, dt

\; I = \frac{2}{a + 1} \sqrt{\frac{a + 1}{a - 1}} \int\; \frac{1}{1 - t^2} \, dt

\; I = \frac{2}{\sqrt{a^2 - 1}} \times \frac{1}{2} \log \left| \frac{1 + t}{1 - t} \right| + C

\; = \frac{1}{\sqrt{a^2 - 1}} \log \left| \frac{1 + \sqrt{\frac{a - 1}{a + 1}} \tan \frac{x}{2}}{1 - \sqrt{\frac{a - 1}{a + 1}} \tan \frac{x}{2}} \right|

\; = \frac{1}{\sqrt{a^2 - 1}}

\; \log \left| \frac{\sqrt{a + 1} \cos \frac{x}{2} + \sqrt{a - 1} \sin \frac{x}{2}}{\sqrt{a + 1} \cos \frac{x}{2} - \sqrt{a - 1} \sin \frac{x}{2}} \right| + C

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