Advertisements
Advertisements
Question
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
Sum
Advertisements
Solution
` Note:" Here, we are considering "log x as log_e x `
\[\text{Let I} = \int\frac{1 + \cot x}{x + \log \sin x}dx\]
\[\text{Putting}\ x + \log \ sin\ x = t\]
\[ \Rightarrow 1 + \ cot\ x = \frac{dt}{dx}\]
\[ \Rightarrow \left( 1 + \cot x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{log} \left| t \right| + C\]
\[ = \text{log }\left| x + \log \sin\ x \right| + C \left[ \because t = x + \log \sin x \right]\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int \left( \tan x + \cot x \right)^2 dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{1}{\left( x + 1 \right)\left( x^2 + 2x + 2 \right)} dx\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{x + 1}{x^2 + x + 3} dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]
\[\int\frac{1}{7 + 5 \cos x} dx =\]
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int \cos^5 x\ dx\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
\[\int\frac{1}{1 + x + x^2 + x^3} \text{ dx }\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
