Mathematics

# Find the integral of $\sqrt { { x }^{ 2 }-{ a }^{ 2 } }$ w.r.t $x$ and hence evaluate $\int { \sqrt { { x }^{ 2 }-8x+7 } } dx\quad$

##### SOLUTION
$I=\int { \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } dx=\sqrt { { x }^{ 2 }-{ a }^{ 2 } } .x-\cfrac { 1 }{ 2 } { \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ 1/2-1 }(2x)$
$=x\sqrt { { x }^{ 2 }-{ a }^{ 2 } } -\int { \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } -{ a }^{ 2 }\int { \cfrac { 1 }{ \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } }$
$\quad I=x\sqrt { { x }^{ 2 }-{ a }^{ 2 } } -1-{ a }^{ 2 }\log { \left| x+\sqrt { { x }^{ 2 }-{ a }^{ 2 } } \right| } \quad$
$I=\int { \sqrt { { x }^{ 2 }-8x+7 } } dx=\int { \sqrt { { \left( x-4 \right) }^{ 2 }-{ 3 }^{ 2 } } } dx=\cfrac { x-4 }{ 2 } \sqrt { { \left( x-4 \right) }^{ 2 }-{ 3 }^{ 2 } } -\cfrac { 9 }{ 2 } \log { \left| \left( x-4 \right) \sqrt { { \left( x-4 \right) }^{ 2 }-{ 3 }^{ 2 } } \right| } +c\quad \quad$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \frac{\sec x}{\log \left ( \sec x+\tan x \right )}dx$
• A. $\displaystyle \log \left [ \log \left ( \sec x-\tan x \right ) \right ].$
• B. $\displaystyle \log \left [ \log \left ( \sin x+\cos x \right ) \right ].$
• C. $\displaystyle \log \left [ \log \left ( \sin x-\cos x \right ) \right ].$
• D. $\displaystyle \log \left [ \log \left ( \sec x+\tan x \right ) \right ].$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int \cos^{-1}\sqrt{\frac{1-x}{2}}dx=$
• A. $\displaystyle \frac{\pi}{2}x-\frac{1}{2}xcos^{-1}x+c$
• B. $\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x-\sqrt{1-x^{2}}+c$
• C. $\displaystyle \dfrac{\pi}{2}x+\dfrac{1}{2}xcos^{-1}x+c$
• D. $\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x+\dfrac{1}{2}\sqrt{1-x^{2}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int(tanx-cotx)^{2}dx=$
• A. $tanx+x+c$
• B. $tanx-x+c$
• C. $tanx-cotx+c$
• D. $tanx-cotx-4x+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\underset{n\rightarrow \infty}{\lim} [\displaystyle \frac{1}{\sqrt{n^{2}-1^{2}}}+\frac{1}{\sqrt{n^{2}-2^{2}}}+\ldots+\frac{1}{\sqrt{2n-1}}]=$
• A. $\pi$
• B. $2\pi$
• C. $3\pi/2$
• D. $\pi/2$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$