CBSE > Class 12 > Mathematics | CBSE Board | Homework Help

Put 1 as sin2x/2 + cos2x/2 and sinx as 2sinx/2cosx/2 then it will become (sinx/2+cosx/2) its whole square then the root will get removed and it will become integration of sinx/2+cosx/2 and now integrate it using basic rules

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X^2 cos x dx intregation by parts

Posted by Kapil Pal 6 years, 2 months ago

CBSE > Class 12 > Mathematics

2 answers

Ron Vineetahuja 6 years, 2 months ago

By ILATE

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Aditya Jha 6 years, 2 months ago

Use the ILATE procedure

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(2x lny) dx + (x^2÷y+3y^2) dy=0

Posted by Tanya Khandelwal 6 years, 2 months ago

CBSE > Class 12 > Mathematics

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Integrate the function 1/ √×+a +√×+b

Posted by Rohit Agrawal 6 years, 2 months ago

CBSE > Class 12 > Mathematics

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Ex11.2 Q15

Posted by Lynda Vung 6 years, 2 months ago

CBSE > Class 12 > Mathematics

0 answers

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Show that log(tan theta +cot theta= pylog2

Posted by Vicky Kumar 6 years, 2 months ago

CBSE > Class 12 > Mathematics

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Integration by substitution Dx/ x2 (x4+1)power 3/4

Posted by Aditya Pratap Singh 6 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Sia ? 6 years, 2 months ago

{tex}\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}{/tex}
{tex}\int \frac{1 d x}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}{/tex}
Taking x⁴ common from denominator
{tex}=\int \frac{1 d x}{x^{2}\left(x^{4}\right)^{\frac{3}{4}}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}{/tex}
{tex}=\int \frac{d x}{x^{2}\left(x^{3}\right)\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}{/tex}
{tex}=\int \frac{d x}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}{/tex}
{tex}\text { Let } t=1+\frac{1}{x^{4}}{/tex}
{tex}\frac{d t}{d x}=-\frac{4}{x^{5}}{/tex}
{tex}-\frac{d t}{4}=\frac{d x}{x^{5}}{/tex}
Substituting value of x and dx
{tex}=\frac{-1}{4} \int \frac{d t}{t^{\frac{3}{4}}}{/tex}
{tex}=\frac{-1}{4} \int t^{\frac{-3}{4}} d t{/tex}
{tex}=\frac{-1}{4} \int t^{\frac{-3}{4}} d t{/tex}
{tex}=\frac{-1}{4}\left[\frac{t^{\frac{-3}{4}}+1}{\frac{-3}{4}+1}\right]+C{/tex}
{tex}=-t^{\frac{1}{4}}+C {/tex}
{tex}=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c{/tex}

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Integrate by using trigonometric Identities 1-cos x/1+cos x dx

Posted by Aditya Pratap Singh 6 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Miss Mor 6 years, 2 months ago

1-cosx=2sin^2x/2 1+cosx=2cos^2x/2 Integration 1-cosx/1+cosx dx =2tan^2x/2dx =2(sec^2x/2-1)dx =2Tanx/2×1/2-2x+c =tanx/2-2x+c

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Miscellaneous exercise on 7 question no. 3

Posted by Anand Kumar 6 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Ron Vineetahuja 6 years, 2 months ago

Which chapter

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Integration of sin(x-c)/sin(x+c)dx

Posted by Khushi Dubey 6 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Mahender Singh 6 years, 2 months ago

Sin (x-c+c-c)=Sin (x+c-2c) After Sin (x-y) using formula

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Integration of xsin²x

Posted by Sushil Kumar 6 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Om Shinde 6 years, 2 months ago

As per the ILATE rule x should be taken as 1st function and (sinx)^2 as second funtion. Then apply the formula of by parts and solve it. The final answer is :-) x^2/4 - xsin2x/4 - cos2x/8

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Integration of x2 + 3 / x4 +9

Posted by Golden Kumar 6 years, 2 months ago

CBSE > Class 12 > Mathematics

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Integration √tanx

Posted by Sunil Kumar 6 years, 2 months ago

CBSE > Class 12 > Mathematics

3 answers

Harshvardhan Gupta 6 years, 2 months ago

Take first function as tanx and second as 1 Then apply by parts If any doubt then analyze the by-parts formula and procedure from maths book

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Harshvardhan Gupta 6 years, 2 months ago

By by-parts 1.√tanx Take first function tanx and function 1 √tanx.integration of 1-integration(derivative of tanx.integration of 1) Then after the derivation substitute tanx=t^2 Then again apply by parts to solve tan inverse t^2

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Aditya Kumar 6 years, 2 months ago

-log|cosx+C| or log|secx+C|

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Integration of tan3x

Posted by Anurag Lodhi 6 years, 2 months ago

CBSE > Class 12 > Mathematics