मराठी

∫ Sin 3 X − Cos 3 X Sin 2 X Cos 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
बेरीज

उत्तर

\[\int\left( \frac{\sin^3 x - \cos^3 x}{\sin^2 x \cdot \cos^2 x} \right)dx\]
\[ = \int\frac{\sin^3 x}{\sin^2 x \cdot \cos^2 x}dx - \int\frac{\cos^3 x}{\sin^2 x \cdot \cos^2 x}dx\]
\[ = \int\frac{\sin x}{\cos^2 x}dx - \int\frac{\cos x}{\sin^2 x}dx\]
\[ = \int\frac{\sin x}{\cos x} \times \frac{1}{\cos x}dx - \int\frac{\cos x}{\sin x} \times \frac{1}{\sin x}dx\]
`=∫ sec x  tan x  dx - ∫  "cosec"  x  cot x  dx`
\[ = \sec x - \left( - \text{cosec  x} \right) + C\]
\[ = \sec x + \text{cosec x }+ C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.02 | Q 23 | पृष्ठ १५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]

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


\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

`  ∫  sin 4x cos  7x  dx  `

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


\[\int\frac{a}{b + c e^x} dx\]

\[\int x^3 \sin x^4 dx\]

\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]

\[\int\frac{x^2}{\sqrt{1 - x}} dx\]

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

`  ∫ \sqrt{"cosec x"- 1}  dx `

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

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

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

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

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

\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]

\[\int\frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \text{ dx }\]

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

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

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

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

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

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

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

\[\int\left( x + 1 \right) \sqrt{2 x^2 + 3} \text{ dx}\]

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

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

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

Evaluate the following integral:

\[\int\frac{x^2}{1 - x^4}dx\]

\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]

\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

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

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

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


\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]


\[\int \sin^3  \left( 2x + 1 \right)  \text{dx}\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×