Advertisements
Advertisements
प्रश्न
उत्तर
\[\int \frac{dx}{1 - \sin\left( \frac{x}{2} \right)}\]
\[ = \int\frac{\left( 1 + \sin \frac{x}{2} \right)}{\left( 1 - \sin \frac{x}{2} \right) \left( 1 + \sin \frac{x}{2} \right)} dx\]
\[ = \int\left( \frac{1 + \sin \frac{x}{2}}{1 - \sin^2 \frac{x}{2}} \right)dx\]
\[ = \int\left( \frac{1 + \sin\frac{x}{2}}{\cos^2 \frac{x}{2}} \right) dx\]
\[ = \int\left( \sec^2 \frac{x}{2} + \sec \frac{x}{2} \text{tan }\frac{x}{2} \right)dx\]
\[ = \frac{\tan \left( \frac{x}{2} \right)}{\frac{1}{2}} + \frac{\sec \left( \frac{x}{2} \right)}{\frac{1}{2}} + C\]
\[ = 2 \left( \tan \frac{x}{2} + \sec \frac{x}{2} \right) + C\]
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)
` ∫ 1/ {1+ cos 3x} ` dx
` = ∫1/{sin^3 x cos^ 2x} dx`
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
