Advertisements
Advertisements
Question
\[\int\frac{\cos x}{1 + \cos x} dx\]
Sum
Solution
\[\int\frac{\cos x}{1 + \cos x}dx\]
\[ = \int\frac{\cos x\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}dx\]
\[ = \int\frac{\cos x - \cos^2 x}{1 - \cos^2 x}dx\]
\[ = \int\frac{\cos x - \cos^2 x}{\sin^2 x}dx\]
\[ = \int\frac{\cos x}{\sin^2 x} - \frac{\cos^2 x}{\sin^2 x}dx\]
\[ = \int\left( \text{cot x cosec x} - \cot^2 x \right)dx\]
\[ = \int\left( \text{cot x cosec x} - cosec^2 x + 1 \right)dx\]
\[ = \int\text{cot x cosec x dx} - \ ∫ co \sec^2 x dx + \int1dx\]
\[ =\text{ - cosec x }+ \cot x + x + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]
\[\int \left( \tan x + \cot x \right)^2 dx\]
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
Integrate the following integrals:
\[\int\text{sin 2x sin 4x sin 6x dx} \]
\[\int\sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\int x^2 \sqrt{x + 2} \text{ dx }\]
` ∫ tan^3 x sec^2 x dx `
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
\[\int\frac{2x}{2 + x - x^2} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int\cos\sqrt{x}\ dx\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int\frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)} dx\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\frac{x + 1}{x^2 + 4x + 5} \text{ dx}\]
\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int\frac{1}{\sec x + cosec x}\text{ dx }\]
\[\int\sqrt{a^2 + x^2} \text{ dx }\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int \log_{10} x\ dx\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]