Evaluate-0-1-1-1-u-4-du-exactly-if-u-coshx-

Question Number 597 by 112358 last updated on 08/Feb/15

E v a l u a t e \int_{0}^{1} \frac{1}{1 + u^{4}} d u e x a c t l y i f

u = c o s h x .

Commented by prakash jain last updated on 08/Feb/15

\cosh x = 0 will give complex values of x .

However the fact u = \cosh x does not

impact value of integral .

Answered by prakash jain last updated on 08/Feb/15

1 + u^{4} = {(u^{2} + 1)}^{2} - 2 u^{2} = (u^{2} + \sqrt{2} u + 1) (u^{2} - \sqrt{2} u + 1)

\frac{1}{1 + u^{4}} = \frac{A u + B}{u^{2} + \sqrt{2} u + 1} + \frac{C u + D}{u^{2} - \sqrt{2} u + 1}

(A + C) u^{3} + (B + D - \sqrt{2} A + \sqrt{2} C) u^{2} + (A + C + D \sqrt{2} - B \sqrt{2}) u

+ (B + D) = 1

A + C = 0 \Rightarrow A = - C \dots . (i)

B + D - A \sqrt{2} + C \sqrt{2} = 0 \dots (ii)

A + C + D \sqrt{2} - B \sqrt{2} = 0 \Rightarrow B = D \dots (iii)

B + D = 1 \dots (iv)

From (i), (ii) and (iv)

1 + C \sqrt{2} + C \sqrt{2} = 0 \Rightarrow C = \frac{- 1}{2 \sqrt{2}} \Rightarrow A = \frac{1}{2 \sqrt{2}}

From (iii) and (iv)

B = \frac{1}{2}, D = \frac{1}{2}

\frac{1}{1 + u^{4}} = \frac{\frac{1}{2 \sqrt{2}} u + \frac{1}{2}}{u^{2} + \sqrt{2} u + 1} + \frac{\frac{- 1}{2 \sqrt{2}} u + \frac{1}{2}}{u^{2} - \sqrt{2} u + 1}

= \frac{1}{2 \sqrt{2}} [\frac{u + \sqrt{2}}{u^{2} + u \sqrt{2} + 1} - \frac{u - \sqrt{2}}{u^{2} - u \sqrt{2} + 1}]

= \frac{1}{4 \sqrt{2}} [\frac{2 u + \sqrt{2}}{u^{2} + u \sqrt{2} + 1} + \frac{\sqrt{2}}{u^{2} + u \sqrt{2} + 1} - \frac{2 u - \sqrt{2}}{u^{2} - u \sqrt{2} + 1} - \frac{\sqrt{2}}{u^{2} - u \sqrt{2} + 1}

\int \frac{2 u + \sqrt{2}}{u^{2} + u \sqrt{2} + 1} d u = \ln (u^{2} + u \sqrt{2} + 1)

\int \frac{2 u + \sqrt{2}}{u^{2} - u \sqrt{2} + 1} d u = \ln (u^{2} - u \sqrt{2} + 1)

\int \frac{1}{u^{2} + u \sqrt{2} + 1} d u = \int \frac{1}{u^{2} + u \sqrt{2} + \frac{1}{2} + \frac{1}{2}} d u

= \int \frac{1}{{(u + \frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2}} d u = \sqrt{2} \arctan \frac{u + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} \arctan (u \sqrt{2} + 1)

\int \frac{1}{u^{2} - u \sqrt{2} + 1} d u = \int \frac{1}{u^{2} - u \sqrt{2} + \frac{1}{2} + \frac{1}{2}} d u =

= \sqrt{2} \arctan (u \sqrt{2} - 1)

Putting all integrals together

\int \frac{1}{1 + u^{4}} d u =

\frac{1}{4 \sqrt{2}} [\ln (u^{2} + u \sqrt{2} + 1) - \ln (u^{2} - u \sqrt{2} + 1) + 2 \arctan (u \sqrt{2} + 1) - 2 \arctan (u \sqrt{2} - 1)]

= \frac{1}{4 \sqrt{2}} [\ln \frac{u^{2} + u \sqrt{2} + 1}{u^{2} - u \sqrt{2} + 1} + 2 \arctan (u \sqrt{2} + 1) - 2 \arctan (u \sqrt{2} - 1)

Value at u = 1

\frac{1}{4 \sqrt{2}} [\ln \frac{2 + \sqrt{2}}{2 - \sqrt{2}} + 2 \arctan (\sqrt{2} + 1) - 2 \arctan (\sqrt{2} - 1)]

Value at u = 0

\frac{1}{4 \sqrt{2}} [\ln 1 + 2 \arctan 1 - 2 \arctan (- 1)] = \frac{π}{4 \sqrt{2}}