Mathematics

# If $\int\left(x^{2010}+x^{\text {804 }}+x^{402}\right)\left(2 x^{160 8}+5 x^{402}+10\right)^{1 / 402} d x$$=\dfrac{1}{10 a}\left(2 x^{2010}+5 x^{804}+10 x^{402}\right)^{a / 402} .$ Then $(a-400)$ is equal to

##### SOLUTION
Let $I=\int\left(x^{2010}+x^{804}+x^{402}\right)\left(2 x^{1608}+5 x^{402}+10\right)^{1 / 402} d x$
$=\int x\left(x^{2009}+x^{508}+x^{401}\right) \cdot\left(2 x^{1608}+5 x^{402}+10\right)^{1 / 402} d x$
$=\int\left(x^{2009}+x^{803}+x^{401}\right) \cdot\left(2 x^{2010}+5 x^{304}+10^{402}\right)^{1 / 402} d x$
Put $2 x^{2010}+5 x^{804}+10^{x^{402}}=t$
$\Rightarrow \quad 4020\left(x^{2009}+x^{803}+x^{401}\right) d x=d t$
$\therefore \quad I=\int \dfrac{1}{4020} \cdot(t)^{1 / 402} d t=\dfrac{1}{4020} \cdot \dfrac{t^{1 / 420+1}}{1 / 402+1}$
$\quad=\dfrac{1}{4020} \cdot \dfrac{t^{403 / 402}}{403 / 402}$
$=\dfrac{1}{4030}\left(2 x^{2010}+5 x^{804}+10^{402}\right)^{403 / 402}$
$\therefore a-400=3$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Subjective Medium
Evaluate: $\displaystyle \int { \dfrac { \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) }{ \left( x-4 \right) \left( x-5 \right) \left( x-6 \right) } dx }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate: $\displaystyle \int^{1}_{0} \dfrac {x\sin^{-1}x}{\sqrt {1 - x^{2}}}dx.$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I = \int \tan^3 x \sec^5 x \: dx$, then I is equal to
• A. $\displaystyle \tan^7 x - \tan^5 x + C$
• B. $\displaystyle \frac {1}{7} \sec^7 x + \frac {1}{5} \sec^5 x + C$
• C. None of these
• D. $\displaystyle \frac {1}{7} \sec^7 x - \frac {1}{5} \sec^5 x + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int\dfrac{\cos x}{\sqrt[3]{\sin^{2}x}}dx=$
• A. $3\sqrt[3]{\sin^{2}x}+c$
• B. $\sqrt[3]{\sin x}+c$
• C. $\sqrt[3]{\sin^{2}x}+c$
• D. $3\sqrt[3]{\sin x}+c$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$