Calculus - Solve the following integrals

1708

$\hspace{-16}\bf{(2):: \int\frac{(5x^4+4x^5)}{(x^5+x+1)^2}dx}$\\\\\\ $\bf{\Rightarrow \int\frac{(5x^4+4x^5)}{x^{10}.(1+x^{-4}+x^{-5})^2}dx}$\\\\\\ $\bf{\Rightarrow \int\frac{\left(5x^{-6}+4x^{-5}\right)}{\left(1+x^{-4}+x^{-5}\right)^2}dx}$\\\\\\ Now Let $\bf{\left(5x^{-6}+4x^{-5}\right)=t}$\\\\\\ Then $\bf{\left(5x^{-6}+4x^{-5}\right)=-dt}$\\\\\\ So $\bf{= -\int \frac{1}{t^2}dt = \frac{1}{t}+\mathbb{C}}$\\\\\\ $\bf{=\frac{1}{1+x^{-4}+x^{-5}}+\mathbb{C}}$\\\\\\ $\bf{=\frac{x^5}{x^5+x+1}+\mathbb{C}}$$

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2305

Shaswata Roy ·Aug 7 '13 at 18:01

$15 . \int \sqrt 1 + \sqrt x 1 - \sqrt x d x = \int \sqrt 1 - x 1 - \sqrt x d x$

Let sin²Î¸ = x

$\int cos θ 1 - sin θ 2 sin θ cos θ d θ$

$= 2 \int sin θ d θ - 2 \int sin 2 θ d θ$

$= - 2 cos θ - 2 \int 2 1 - cos 2 θ d θ$

$= - 2 cos θ - θ + 2 sin 2 θ + C$

$= - 2 \sqrt 1 - x - sin - 1 \sqrt x + 2 \sqrt x \sqrt 1 - x + C$

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1708

man111 singh ·Oct 19 '13 at 22:55

$\hspace{-16}\bf{(6)\;\; \int \left(\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right)\cdot \ln (x)dx}$\\\\\\ Let $\bf{\left(\left(\frac{x}{e}\right)^x-\left(\frac{e}{x}\right)^x\right)=t\;\;,}$ Then \\\\\\ $\bf{\left(\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right)\cdot \ln(x)dx=dt}$\\\\\\ So Integration is $\bf{\int 1\cdot dt = t+\mathbb{C}}$\\\\\\ So $\bf{\int \left(\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right)\cdot \ln (x)dx=\left(\left(\frac{x}{e}\right)^x-\left(\frac{e}{x}\right)^x\right)+\mathbb{C}}$$

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Divyans Jhunjhunwala ·Jul 23 '13 at 9:21

answer for 16...

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Divyans Jhunjhunwala ·Jul 23 '13 at 9:31

answer for 8...

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1357

Manish Shankar ·Jul 25 '13 at 21:08

Are they very easy?

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Harsh Bharvada ·Aug 3 '13 at 2:16

each and every book i look into i get the 15 number sum ......sir u have even given it in shastras..can u plzzz given the solution ?

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Divyans Jhunjhunwala ·Aug 7 '13 at 23:15

Take x= cos²Î¸
thn its coming - log( cosÎ¸)
Sustitute for Î¸ nd u get the ans...

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Divyans Jhunjhunwala For ques 15
Upvote·0· Reply ·Aug 7 '13 at 23:16
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Divyans Jhunjhunwala ·Aug 7 '13 at 23:18

Sir help for question 9...

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Manish Shankar ·Aug 8 '13 at 1:07

For 9
multiply, divide by (a²+b²)

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Anik Chatterjee ·Aug 8 '13 at 2:30

For 12,..
-cosa.sin^-1(cos xcos a)-sina.log|sin x-√sin²x-sin²a|+C
??

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Shaswata Roy ·Aug 8 '13 at 8:29

$13) \int sin - 11 / 3 x \cdot cos - 1 / 3 x d x$
$= \int tan - 11 / 3 x \cdot sec 4 x d x$
$= \int tan - 11 / 3 x (1 + tan 2 x) d tan x$
$= - 8 3 tan - 8 / 3 - 2 3 tan - 2 / 3 x + C$

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1708

man111 singh ·Oct 19 '13 at 23:25

$\hspace{-16}\bf{(11)\;\int\sqrt{x+\sqrt{x^2+2}}\; dx}$\\\\\\ Let $\bf{\boxed{\bf{\left(x+\sqrt{x^2+2}\right)=t^2}}\Leftrightarrow \left(1+\frac{1}{2\sqrt{x^2+2}}.2x\right)dx = 2tdt}$\\\\\\ $\bf{\left(\frac{x+\sqrt{x^2+2}}{\sqrt{x^2+2}}\right)dx =2tdt\Leftrightarrow dx = \frac{2\cdot \sqrt{x^2+2}}{t}dt}$\\\\\\ Now using $\bf{\left(\sqrt{x^2+2}+x\right)\cdot\left(\sqrt{x^2+2}-x\right)=2}$\\\\\\ $\bf{t^2\cdot\left(\sqrt{x^2+2}-x\right)=2\Leftrightarrow \boxed{\bf{\left(\sqrt{x^2+2}-x\right)=\frac{2}{t^2}}}}$\\\\\\ So $\bf{\sqrt{x^2+2}=\frac{1}{2}\cdot \left(t^2+\frac{2}{t^2}\right) = \frac{1}{2}\cdot \left(\frac{t^4+2}{t^2}\right)}$\\\\\\ So Integration is $\bf{\int t\cdot \frac{\left(t^4+2\right)}{t^2}.\frac{1}{t}dt}$\\\\\\ So $\bf{\int \left(\frac{t^4+2}{t^2}\right)dt = \int t^2dt+2\cdot \int t^{-2}dt }$\\\\\\ So $\bf{=\frac{1}{3}\cdot t^3 -\frac{2}{t}+\mathbb{C}}$\\\\\\$

$\hspace{-16}$So $\bf{=\frac{1}{3}\cdot \left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\cdot \left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+\mathbb{C}}$\\\\\\ $\bf{\bullet }$Alternatively..........\\\\\\ Using $\bf{\left(x+\sqrt{x^2+2}\right)=e^{2t}\;\;, }$ Then $\bf{\left(\sqrt{x^2+2}-x\right)=2\cdot e^{-2t}}$\\\\\\ Integral convert into $\bf{\int e^t \cdot \left(e^{2t}+2e^{-2t}\right)dt}$$

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1708

man111 singh ·Oct 20 '13 at 0:06

$\hspace{-16}\bf{(9)\;\; \int\frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}dx}$\\\\\\ Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{\cos^2 x}$ \\\\\\ $\bf{\int\frac{a^2\tan^2 x+b^2}{a^4\tan^2 x+b^4}dx = \frac{1}{a^4}\int \frac{\tan^2 x+\left(\frac{b}{a}\right)^2}{\tan^2 x+\left(\frac{b}{a}\right)^4}dx}$\\\\\\ Now Put $\bf{\left(\frac{b}{a}\right)=d}$ and $\bf{\tan x=t \Leftrightarrow \sec^2 xdx =dt}$\\\\\\ $\bf{dx=\frac{1}{(1+t^2)}dt}$\\\\\\ So Integral is $\bf{\frac{1}{a^4}\int\frac{t^2+d}{(t^2+d^2)\cdot (1+t^2)}dt}$\\\\\\ Now Using Partial fraction \\\\\\ $\bf{=\frac{1}{a^4\cdot(d+1)}\left(\int \frac{d}{(t^2+d^2)}dt+\int\frac{1}{(t^2+1)}dt\right)}$\\\\\\ $\bf{=\frac{1}{a^4}\cdot \tan^{-1}\left(\frac{t}{d}\right)+\frac{1}{a^4\cdot (d+1)}\cdot \tan^{-1}(t)+\mathbb{C}}$\\\\\\ So $\bf{\int\frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}dx=\frac{1}{a^4}\cdot \tan^{-1}\left(\frac{\tan x}{d}\right)+\frac{1}{a^4\cdot (d+1)}\cdot \tan^{-1}(\tan x)+\mathbb{C}}$ \\\\\\$

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Shaswata Roy ·Nov 13 '13 at 9:14

$\int \sqrt sin 4 θ + cos 4 θ - cos 2 θ sin 2 θ tan 2 θ d θ $
$= \int \sqrt tan 4 θ + 1 - tan 2 θ 1 - tan 2 θ 2 tan θ \cdot sec 2 θ d θ $

Substituting tan²Î¸=z,

$I = \int (1 - z) \sqrt z 2 + 1 - z d z $

Substituting 1-z=1t,

$I = \int t 1 \sqrt (1 - t 1 ) 2 + 1 - (1 - t 1 ) t 2 1 d t $
$= \int \sqrt t 2 - 1 + t d t $
$= \int \sqrt (t - 2 1 ) 2 - 4 5 d t $
$= \sqrt 5 1 ln ∣ t - 2 1 + \sqrt t 2 - 1 + t ∣ + C$

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Shaswata Roy Answer for question 1.
Upvote·0· Reply ·Nov 13 '13 at 9:16
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