Advertisements
Advertisements
प्रश्न
उत्तर
\[\int\left( \frac{1 + \cos 4x}{\cot x - \tan x} \right) dx\]
\[ = \int\frac{\left( 1 + \cos 4x \right)}{\left( \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \right)} dx\]
\[ = \int\frac{2 \cos^2 2x \times \sin x \cos x}{\left( \cos^2 x - \sin^2 x \right)}dx\]
\[ = \int\frac{\cos^2 2x \times 2 \sin x \cos x}{\cos 2x}dx\]
\[ = \int\cos 2x \sin 2xdx\]
\[ = \frac{1}{2}\int2 \sin 2x \cos 2xdx\]
\[ = \frac{1}{2}\int\sin 4xdx\]
\[ = \frac{1}{2}\left[ - \frac{\cos 4x}{4} \right] + C\]
\[ = - \frac{1}{8}\cos 4x + C\]
APPEARS IN
संबंधित प्रश्न
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
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
