मराठी

∫ Cos √ X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\cos\sqrt{x}\ dx\]
बेरीज

उत्तर

\[\text{ Let I } = \int\cos \sqrt{x} dx\]
\[ = \int\frac{\sqrt{x} \cdot \cos \sqrt{x}}{\sqrt{x}}dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[ \therefore I = 2\int t_{} \cdot \cos \left( t \right)_{} \cdot dt\]
\[\text{Taking t as the first function and cos t as the second function} . \]
\[ = 2 \left[ t \cdot \sin t - \int1 \cdot \text{ sin t dt }\right]\]
\[ = 2 \left[ t \cdot \sin t + \cos t \right] + C . . . . (1) \]
\[\text{Substituting the value of t in eq} \text{ (1) }\]
\[ = 2 \left[ \sqrt{x} \cdot \sin \sqrt{x} + \cos \sqrt{x} \right] + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.25 [पृष्ठ १३३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.25 | Q 26 | पृष्ठ १३३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]

\[\int \left( e^x + 1 \right)^2 e^x dx\]

\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]

\[\int\frac{1 - \cot x}{1 + \cot x} dx\]

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


\[\int\frac{1 - \sin x}{x + \cos x} dx\]

\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]

\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]

\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]

\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]

\[\int \cot^5 x  \text{ dx }\]

\[\int \sin^4 x \cos^3 x \text{ dx }\]

Evaluate the following integrals:

\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{  dx }\]

\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]

\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]

\[\int\frac{1}{3 + 4 \cot x} dx\]

\[\int x \cos x\ dx\]

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

\[\int x \sin x \cos 2x\ dx\]

\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]

∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]

\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]

\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]

\[\int\left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]

\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]

Write a value of

\[\int e^{3 \text{ log x}} x^4\text{ dx}\]

\[\int\frac{1}{1 + \tan x} dx =\]

\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]

\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to


\[\int \text{cosec}^2 x \text{ cos}^2 \text{  2x  dx} \]

\[\int\sin x \sin 2x \text{ sin  3x  dx }\]


\[\int \cos^5 x\ dx\]

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

\[\int\sqrt{3 x^2 + 4x + 1}\text{  dx }\]

\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

\[\int \cos^{- 1} \left( 1 - 2 x^2 \right) \text{ dx }\]

Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×