integrals for zzz and others

1.

x² = tan θ

integral reduces too

12∫√cotθ dθ

which is standard

\frac{1}{2}\int \frac{(x^2+1)+(1-x^2)}{1+x^4}
now u can go for algebraic twins ;)

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 .

q2 right substituition

3.

on substitutin

x^( 2/3)= t

we get

∫ √t+ t√t1+ t√t dt

whch can be done

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