Advertisements
Advertisements
Question
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
Sum
Advertisements
Solution
\[\int \tan^\frac{3}{2} x \cdot \sec^2 \text{x dx}\]
\[Let \tan x = t\]
\[ \Rightarrow \sec^2 x = \frac{dt}{dx}\]
\[ \Rightarrow \sec^2 \text{x dx} = dt\]
\[Now, \int \tan^\frac{3}{2} x \cdot \sec^2 \text{x dx} \]
\[ = \int t^\frac{3}{2} dt\]
` = [t^{3/2 +1}/{3/2+1}] + C`
\[ = \frac{2}{5} t^\frac{5}{2} + C\]
\[ = \frac{2}{5} \tan^\frac{5}{2} x + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{x + 1}{x^2 + x + 3} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int x \text{ sin 2x dx }\]
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ dx }\]
\[\int\left( x + 1 \right) \text{ log x dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int \sin^3 \sqrt{x}\ dx\]
\[\int\frac{x^2 + 1}{\left( 2x + 1 \right) \left( x^2 - 1 \right)} dx\]
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\frac{\log \left( \log x \right)}{x} \text{ dx}\]
\[\int\frac{\log x}{x^3} \text{ dx }\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
