\frac{1}{2}\int \frac{(x^2+1)+(1-x^2)}{1+x^4}
now u can go for algebraic twins ;)
\boxed{1} \int \frac{\mathrm{dx}}{1+x^4} \\ \boxed{2}\int \frac{\sin x+\sin 2x}{\sqrt{\cos x+\cos 2x}}\mathrm{dx}\\ \boxed{3}\int \frac{1+x^{\frac{2}{3}}}{1+x}\mathrm{dx}
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11 Answers
I 1 = ∫ x 2 + 1x 4 + 1 dx
Let ' s put ------ x - 1x = z ;
So , ( 1 + 1x 2 ) dx = dz
I 1 = ∫ dxz 2 + 2 = 1√2 tan - 1 z√2 + c = 1√2 tan - 1 { x 2 - 1√2 x } + c
I 2 = ∫ x 2 - 1x 4 + 1 dx
Similarly proceeding ( Taking x + 1x = z ) and integrating , we get ,
I 2 = 12 √2 log { x 2 - √2 x + 1x 2 + √2 x + 1 } + c '
Clearly , I = ∫ dx1 + x 4 = 12 ( I 1 - I 2 )
Hence , the answer .
Q 2 . Easier method - ( If I am correct in calculations ) -
I = ∫ ( sin x + sin 2 x ) dx√cos x + cos 2x
= ∫ sin x ( 2 cos x + 1 ) dx√ 2 cos 2 x + cos x - 1
= √2 ∫ sin x ( 2 cos x + 1 ) dx√ 4 cos 2 x + 2 cos x - 2
= - √22 ∫ ( 2 cos x + 1 / 2 + 1 / 2 ) ( - 2 sin x dx )√ ( 2 cos x + 1 /2 ) 2 - 9 / 4
Now substitute , 2 cos x + 1 / 2 = z = > - 2 sin x dx = dz
So , I = - 1√2 ∫ ( z + 1 / 2 ) dz√ z 2 - ( 3 / 2 ) 2
I shall take 94 = a 2 for my convenience .
I = - 1√2 ∫ z dz√ z 2 - a 2 - 12 √2 ∫ dz√ z 2 - a 2
= - 1√2 √ z 2 - a 2 - 12 √2 log { z + √z 2 - a 2 } + c
= - √ 2 cos 2 x + cos x - 1 - 12 √2 log { 2 cos x + 1 / 2 + √ 4 cos 2 x + 2 cos x - 2 } + c
zzz one suggestion please finish of a problem before giving suggestions lik "which can be done " or " which is standard "
its good for u only as it will give a feel of finishing a problem [1]
ricky u r right