Question

please guys answer this question please please guys please.........​

Answer 1

AnswEr :

Given Expression,

$\displaystyle \sf \: l = \int {tan}^{ - 1} \bigg( \sqrt{ \dfrac{1 - x}{1 + x} } \bigg)dx$

Let x = cos(y)

We know that,

cos2∅ = cos²∅ - sin²∅

Therefore,

cos2∅ = 2cos²∅ - 1 or, 1 - 2sin²∅

Implies,

• 1 - cos2∅ = 2sin²∅

• 1 + cos2∅ = 2cos²∅

Now,

$\dashrightarrow \displaystyle \sf \: l = \int {tan}^{ - 1} \bigg( \sqrt{ \dfrac{1 - cos(y)}{1 + cos(y)} } \bigg)dx \\ \\ \displaystyle \dashrightarrow \sf \: l = \int {tan}^{ - 1} \bigg( \sqrt{ \dfrac{ \cancel{2} {sin}^{2}( \frac{y}{2} )}{ \cancel{2} {cos}^{2}( \frac{y}{2}) } } \bigg)dx \\ \\ \displaystyle \dashrightarrow \sf \: l = \int {tan}^{ - 1} \bigg( \sqrt{tan {}^{2}( \frac{y}{2}) } \bigg)dx \\ \\ \displaystyle \dashrightarrow \sf \: l = \int {tan}^{ - 1} \bigg(tan {}^{}( \frac{y}{2}) \bigg)dx \\ \\ \displaystyle \dashrightarrow \sf \: l = \dfrac{1}{2} \int \: y.dx$

When x = cos(y), y would be arccos(x)

Thus,

$\displaystyle \dashrightarrow \sf \: l = \dfrac{1}{2} \int cos {}^{ - 1}x .dx \\ \\ \displaystyle \dashrightarrow \sf \: l = \dfrac{1}{2} \times \bigg( \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } } \bigg) \\ \\ \displaystyle \dashrightarrow \boxed{ \boxed{\sf \: l = \dfrac{ - 1}{2 \sqrt{1 - {x}^{2} } } + c }}$