Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{ Let I } = \int \text{ tan}^4 \text{ x dx }\]
\[ = \int \tan^2 x \cdot \tan^2 \text{ x dx}\]
\[ = \int\left( \sec^2 x - 1 \right) \tan^2 \text{ x dx}\]
\[ = \int \sec^2 x \cdot \tan^2\text{ x dx }- \int \tan^2 \text{ x dx}\]
\[ = \int \tan^2 x \cdot \sec^2 x - \int\left( \sec^2 x - 1 \right) dx\]
\[\text{ Putting tan x } = \text{ t in the Ist integral}\]
\[ \Rightarrow \sec^2 \text{ x dx } = dt\]
\[ \therefore I = \int t^2 \cdot dt - \int\left( \sec^2 x - 1 \right) dx\]
\[ = \frac{t^3}{3} - \tan x + x + C\]
\[ = \frac{\tan^3 x}{3} - \text{ tan x + x + C }..............\left( \because t = \tan x \right)\]
APPEARS IN
संबंधित प्रश्न
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .