Advertisements
Advertisements
Question
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
Sum
Solution
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right)dx\]
` = ∫ cot ^-1 (( 2 sin x cos x) /( 2 sin^2 x))` dx ` [∴ sin 2x = 2 sin x cos x & 1 - cos 2x = 2 sin^2 x ]`
\[ = \int \cot^{- 1} \left( \cot x \right)dx\]
` = ∫ x dx `
\[ = \frac{x^2}{2} + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
\[\int\frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} dx\]
\[\int \tan^2 \left( 2x - 3 \right) dx\]
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int \cos^2 \text{nx dx}\]
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int \cot^6 x \text{ dx }\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int\frac{1}{2 x^2 - x - 1} dx\]
\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int x^2 \sin^2 x\ dx\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{x^3}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\frac{1}{1 - \cos x - \sin x} dx =\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int \tan^5 x\ dx\]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\sqrt{a^2 - x^2}\text{ dx }\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int \tan^{- 1} \sqrt{x}\ dx\]
