Advertisements
Advertisements
Question
` ∫ tan^5 x sec ^4 x dx `
Sum
Solution
` ∫ tan^5 x sec ^4 x dx `
= ∫ tan5 x. sec2 x . sec2 x dx
= ∫ tan5 x (1 + tan2 x) sec2 x dx
Let tan x = t
⇒ sec2 x dx = dt
Now, ∫tan5x (1+tan2 x) sec2 x dx
= ∫ t5 (1 + t2) dt
= ∫ (t5 + t7) dt
\[= \frac{t^6}{6} + \frac{t^8}{8} + C\]
\[ = \frac{\tan^6 x}{6} + \frac{\tan^8 x}{8} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{1}{1 - \cos x} dx\]
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{a}{b + c e^x} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
\[\int \cot^5 x \text{ dx }\]
\[\int \cos^5 x \text{ dx }\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int x^2 \text{ cos x dx }\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{x}{\left( x - 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{1 + x + x^2 + x^3} dx\]
\[\int\frac{1}{x \left( x^4 + 1 \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
\[\int x\sqrt{2x + 3} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]
\[\int \log_{10} x\ dx\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]
