Advertisements
Advertisements
Question
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
Sum
Advertisements
Solution
` ∫ cos x / \sqrt{4-sin^2 x}`
\[\text{ let }\sin x = t\]
\[ \Rightarrow \text{ cos x dx }= dt\]
Now, ` ∫ cos x / \sqrt{4-sin^2 x}`
\[ = \int\frac{dt}{\sqrt{4 - t^2}}\]
\[ = \int\frac{dt}{\sqrt{2^2 - t^2}}\]
\[ = \sin^{- 1} \left( \frac{t}{2} \right) + C\]
\[ = \sin^{- 1} \left( \frac{\sin x}{2} \right) + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
\[\int\frac{1}{1 - \cos 2x} dx\]
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx\]
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
\[\int \sin^5 x \text{ dx }\]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{1}{a^2 x^2 + b^2} dx\]
\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int x \text{ sin 2x dx }\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \sqrt{x + 2}}\text{ dx}\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int \sec^6 x\ dx\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
