Find-the-value-of-Q-if-0-Q-cosec-1-d-ln-3-2-2-2-

Question Number 167922 by peter frank last updated on 29/Mar/22

Find the value of Q if

\int_{0}^{Q} \sqrt{cosec θ - 1} d θ = \ln (\frac{3 + 2 \sqrt{2}}{2})

Commented by MJS_new last updated on 29/Mar/22

d x ?

Commented by peter frank last updated on 29/Mar/22

d θ

Answered by MJS_new last updated on 29/Mar/22

\int \sqrt{\csc θ - 1} d θ =

[t = \sqrt{\csc θ - 1} \to d θ = - \frac{2 d t}{(t^{2} + 1) \sqrt{t^{2} + 2}}]

= - 2 \int \frac{t}{(t^{2} + 1) \sqrt{t^{2} + 2}} =

[u = \frac{t + \sqrt{t^{2} + 2}}{\sqrt{2}} \to d t = \frac{\sqrt{t^{2} + 2}}{u}]

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

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

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

\dots

= \ln (1 + 2 \sin x + 2 \sqrt{(1 + \sin x) \sin x}) + C

this equals 0 with x = 0

\Rightarrow

1 + 2 \sin Q + 2 \sqrt{(1 + \sin Q) \sin Q} = \frac{3 + 2 \sqrt{2}}{2}

\Rightarrow \sin Q = \frac{11 - 6 \sqrt{2}}{8}

\dots

Answered by peter frank last updated on 29/Mar/22

\int_{0}^{Q} \sqrt{cosec θ - 1} d θ = \int_{0}^{Q} \sqrt{\frac{1 - \sin θ}{\sin θ}} d θ

\int_{0}^{Q} \sqrt{\frac{1 - \sin θ}{\sin θ}} d θ = \ln (\frac{3 + 2 \sqrt{2}}{2})

\dots \dots .

Answered by peter frank last updated on 30/Mar/22

Answered by peter frank last updated on 30/Mar/22

Commented by peter frank last updated on 30/Mar/22

thank you Mjs for you idea