Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\text{ Putting x }= \cos \theta\]
\[ \Rightarrow dx = - \sin \text{ θ dθ}\]
\[ \therefore I = \int \tan^{- 1} \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \left( - \sin \text{ θ dθ} \right)\]
\[ = \int \tan^{- 1} \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} \left( - \sin \theta \right)d\theta\]
\[ = \int \tan^{- 1} \left( \tan \frac{\theta}{2} \right) \left( - \sin \theta \right)d\theta\]
\[ = - \frac{1}{2}\int\theta \sin \theta d\theta\]
\[\text{Considering} \text{ θ as first function and} \sin \text{ θ as second function}\]
\[I = - \frac{1}{2}\left[ \theta\left( - \cos \theta \right) - \int1\left( - \cos \theta \right)d\theta \right]\]
\[ = - \frac{1}{2}\left( \theta\left( - \cos \theta \right) + \int\cos \theta d\theta \right)\]
\[ = - \frac{1}{2}\left( - \theta \cos \theta + \sin \theta \right) + C\]
\[ = - \frac{1}{2}\left[ - \theta \cos \theta + \sqrt{1 - \cos^2 \theta} \right] + C\]
\[ = - \frac{1}{2}\left[ - \cos^{- 1} x \times x + \sqrt{1 - x^2} \right] + C\]
\[ = \frac{1}{2}\left[ x \cos^{- 1} x - \sqrt{1 - x^2} \right] + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
` ∫ tan x sec^4 x dx `
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
