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
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{1}{1 - \sin x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
` ∫ sin 4x cos 7x dx `
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
\[\int\frac{dx}{e^x + e^{- x}}\]
\[\int\frac{1}{\sqrt{5 x^2 - 2x}} dx\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{ dx }\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int\cos\sqrt{x}\ dx\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{1}{x \left( x^4 + 1 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
Find \[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)^2}dx\]
\[\int\sqrt{\cot \text{θ} d } \text{ θ}\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int\sqrt{x^2 - a^2} \text{ dx}\]
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
\[\int x^2 \tan^{- 1} x\ dx\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
