Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[We\ know\ that, \]
\[ \tan 5x = \tan \left( 2x + 3x \right)\]
\[ \Rightarrow \tan 5x = \frac{\tan 2x + \tan 3x}{1 - \tan 2x \tan 3x}\]
\[ \Rightarrow \tan 5x - \tan 2x \tan 3x \tan 5x = \tan 2x + \tan 3x\]
\[ \Rightarrow \tan 2x \tan 3x \tan 5x = \tan 5x - \tan 2x - \tan 3x\]
\[ \therefore \int\tan 2x \tan 3x \tan 5x = \int\left( \tan 5x - \tan 2x - \tan 3x \right)dx\]
\[ = \frac{1}{5} \ln \left| \sec 5x \right| - \frac{1}{2} \ln \left| \sec 2x \right| - \frac{1}{3} \ln \left| \sec 3x \right| + C\]
APPEARS IN
संबंधित प्रश्न
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
