Advertisements
Advertisements
Question
` ∫ sec^6 x tan x dx `
Sum
Advertisements
Solution
` ∫ sec^6 x tan x dx `
=∫ sec6 x.sec x tan x dx
Let sec x = t
Let sec x = t
⇒ sec x tan x dx = dt
Now, ∫ sec6 x.sec x tan x dx
= ∫ t6. dt
Now, ∫ sec6 x.sec x tan x dx
= ∫ t6. dt
\[= \frac{t^6}{6} + C\]
\[ = \frac{\sec^6 x}{6} + C\]
\[ = \frac{\sec^6 x}{6} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
`∫ cos ^4 2x dx `
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
` ∫ tan^3 x sec^2 x dx `
\[\int \sin^4 x \cos^3 x \text{ dx }\]
Evaluate the following integrals:
\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]
\[\int\frac{1}{a^2 x^2 + b^2} dx\]
\[\int\frac{1}{x^2 + 6x + 13} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]
\[\int\frac{1}{2 + \sin x + \cos x} \text{ dx }\]
\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2} \text{ dx }\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]
\[\int\frac{1}{e^x + e^{- x}} dx\]
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[\int\log \left( x + \sqrt{x^2 + a^2} \right) \text{ dx}\]
\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
