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DIVYA SHARMA Grade: 12th Pass
        


1.∫(e200x+e202x)/(ex+e-x)dx.


2.∫ex[(2-sin2x)/(1-cos2x)]dx


3.∫[x5]/[(x2+x+1)(x6+1)(x4-x3+x-1)]dx


4.∫[x3+2x2+x+2]/[(x2+1)1/2]dx


5.∫dx/(5-13sinx)


6.∫dx/(a2-b2cos2x)    a2>b2


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 

5 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										
Ans:
I = \int \frac{e^{200x} + e^{202x}}{e^{x}+e^{-x}}dx
I = \int e^{201x}dx = \frac{e^{201x}}{201} + constant
I = \int e^{x}.\frac{2-sin2x}{1-cos2x}dx
I = \int e^{x}.\frac{2-2sinx.cosx}{2sin^{2}x}dx
I = \int e^{x}.(csc^{2}x - cotx)dx
I = \int -e^{x}.(cotx-csc^{2}x)dx
I = -\int (e^{x}.cotx)'dx = -e^{x}.cotx + constant
I = \int \frac{x^{5}}{(x^{2}+x+1)(x^{6}+1)(x^{4}-x^{3}+x^{2}-1)}
Simply using the partial fraction here, we have
I = \frac{1}{12}(log(\frac{1-x^{6}}{1+x^{6}})) + constant
I = \frac{1}{6}tanh^{-1}(x^{6}) + constant
I = \int \frac{x^{3}+2x^{2}+x+1}{\sqrt{x^{2}+1}}dx
t = tanx
dt = sec^{2}x.dx
I = \int (sect+tan^{3}t.sect+2tan^{2}t.sect+sect.tant)dt
I = \frac{sec^{3}t}{3}+sect.tant + constant
I = \frac{1}{3}\sqrt{x^{2}+1}(x^{2}+3x+1)+constant
I = \int \frac{1}{5-13sinx}dx
t = tan(\frac{x}{2})
dt = \frac{1}{2}sec^{2}(\frac{x}{2})dx
I = 2\int \frac{1}{5t^{2}-26t+5}dt
I = \frac{1}{6}tanh^{-1}(\frac{1}{12}(13-5t)) + constant
I = \frac{1}{6}tanh^{-1}(\frac{1}{12}(13-5tan(\frac{x}{2}))) + constant
I = \int \frac{1}{a^{2}-b^{2}cos^{2}x}dx
I = \int \frac{sec^{2}x}{a^{2}sec^{2}x-b^{2}}dx
I = \int \frac{sec^{2}x}{a^{2}+a^{2}tan^{2}x-b^{2}}dx
t = tanx
dt = sec^{2}xdx
I = \int \frac{1}{a^{2}+a^{2}t^{2}-b^{2}}dt
I = \frac{1}{a^{2}-b^{2}}\int \frac{1}{\frac{a^{2}t^{2}}{a^{2}-b^{2}}+1}dt
I = \frac{tan^{-1}(\frac{at}{\sqrt{a^{2}-b^{2}}})}{a\sqrt{a^{2}-b^{2}}} + constant
I = \frac{tan^{-1}(\frac{atanx}{\sqrt{a^{2}-b^{2}}})}{a\sqrt{a^{2}-b^{2}}} + constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
3 years ago
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