Advertisements
Advertisements
Question
\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]
Sum
Advertisements
Solution
\[\int\frac{\cos \left( 2 x \right) \cdot dx}{\sqrt{\sin^2 2x + 8}}\]
\[\text{ let } \text{ sin } \left( 2x \right) = t\]
\[ \Rightarrow \text{ cos }\left( 2x \right) \times 2 \cdot dx = dt\]
\[ \Rightarrow \text{ cos }\left( 2x \right) \cdot dx = \frac{dt}{2}\]
\[Now, \int\frac{\text{ cos } \left( 2 x \right) \cdot dx}{\sqrt{\sin^2 2x + 8}} \]
\[ = \frac{1}{2}\int\frac{dt}{\sqrt{t^2 + \left( 2\sqrt{2} \right)^2}}\]
\[ = \frac{1}{2}\text{ log }\left| t + \sqrt{t^2 + 8} \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| \text{ sin }\left( 2x \right) + \sqrt{\text{ sin }^2 \left(\text{ 2x }\right) + 8} \right| + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int \left( 3x + 4 \right)^2 dx\]
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\frac{x + 3}{\left( x + 1 \right)^4} dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int x^3 \cos x^4 dx\]
\[\int x^3 \sin x^4 dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int\frac{1}{1 + x - x^2} \text{ dx }\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{2x}{2 + x - x^2} \text{ dx }\]
\[\int\frac{x + 2}{2 x^2 + 6x + 5}\text{ dx }\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{\cos 2x + 3 \sin^2 x} dx\]
\[\int\frac{1}{2 + \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int x \text{ sin 2x dx }\]
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
\[\int\sqrt{2ax - x^2} \text{ dx}\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int \cos^5 x\ dx\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
