English

∫ 1 √ 8 + 3 X − X 2 D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int\frac{dx}{\sqrt{8 + 3x - x^2}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x^2 - 3x \right)}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x^2 - 3x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 \right)}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{8 - \left( x - \frac{3}{2} \right)^2 + \frac{9}{4}}}\]
\[ \Rightarrow \int\frac{dx}{\sqrt{\left( \frac{\sqrt{41}}{2} \right)^2 - \left( x - \frac{3}{2} \right)^2}}\]
\[ \Rightarrow \sin^{- 1} \left( \frac{x - \frac{3}{2}}{\frac{\sqrt{41}}{2}} \right) + C\]
\[ \Rightarrow \sin^{- 1} \left( \frac{2x - 3}{\sqrt{41}} \right) + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.17 [Page 93]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.17 | Q 2 | Page 93

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]

`  ∫  sin 4x cos  7x  dx  `

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

\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]

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

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

\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

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

Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]

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

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

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

`  ∫ \sqrt{"cosec x"- 1}  dx `

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

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

\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]

\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]

`int 1/(sin x - sqrt3 cos x) dx`

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

\[\int\frac{1}{p + q \tan x} \text{ dx  }\]

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

\[\int\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} \text{ dx }\]

\[\int x e^{2x} \text{ dx }\]

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

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

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

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

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

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

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

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

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×