Advertisements
Advertisements
Question
\[\int\cos\sqrt{x}\ dx\]
Sum
Advertisements
Solution
\[\text{ Let I } = \int\cos \sqrt{x} dx\]
\[ = \int\frac{\sqrt{x} \cdot \cos \sqrt{x}}{\sqrt{x}}dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[ \therefore I = 2\int t_{} \cdot \cos \left( t \right)_{} \cdot dt\]
\[\text{Taking t as the first function and cos t as the second function} . \]
\[ = 2 \left[ t \cdot \sin t - \int1 \cdot \text{ sin t dt }\right]\]
\[ = 2 \left[ t \cdot \sin t + \cos t \right] + C . . . . (1) \]
\[\text{Substituting the value of t in eq} \text{ (1) }\]
\[ = 2 \left[ \sqrt{x} \cdot \sin \sqrt{x} + \cos \sqrt{x} \right] + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\left( x^e + e^x + e^e \right) dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]
\[\int\frac{x^2}{\sqrt{x - 1}} dx\]
` ∫ { x^2 dx}/{x^6 - a^6} dx `
\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int2 x^3 e^{x^2} dx\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int \cos^3 (3x)\ dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\]
\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]
\[\int \tan^3 x\ dx\]
\[\int \tan^5 x\ dx\]
\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
