Question

Integrate the following :-
$\sf{\int\limits^\frac{\pi}{2}_0 {\dfrac{sin^2(x)}{sin(x)cos(x)}} \, dx }$

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Answer 1

GIVEN EXPRESSION :-

$\sf{\int\limits^\frac{\pi}{2}_0 {\dfrac{sin^2(x)}{sin(x)cos(x)}} \, dx }$

OBJECTIVE :-

To integrate the given expression.

SOLUTION :-

By tangent half-angle* substitution :

$\sf{=\int \dfrac{4tan^2\frac{x}{2}}{[tan^2(\frac{x}{2})+1]^2 \bigg[\dfrac{1-tan^2(\frac{x}{2})}{tan^2(\frac{x}{2})+1}+\dfrac{2tan(\frac{x}{2})}{tan^2(\frac{x}{2})+1}\bigg]}}$

$\rule{150}{1}$

$\sf{Let\ us\ substitute\ tan(\frac{x}{2})=u.}\\\\\sf{So,\ on\ differentiating\ it\ we\ get\ \dfrac{du}{dx}=\dfrac{sec^2(\frac{x}{2})}{2}}.$

$\sf{\longrightarrow dx=\dfrac{2}{sec^2(\frac{x}{2})}du}\\\\\\\sf{\longrightarrow dx=\dfrac{2}{u^2+1}du}$

$\rule{150}{1}$

$\sf{=-8\int \dfrac{u^2}{(u^2+1)^2(u^2-2u-1)}du}$

$\rule{150}{1}$

$\sf{Again\ solving\ \int \dfrac{u^2}{(u^2+1)^2(u^2-2u-1)}du}$

By partial fraction decomposition,

$\sf{=\int \bigg(\dfrac{1}{8(u^2-2u-u)}-\dfrac{1}{8(u^2+1)}-\dfrac{u-1}{4(u^2+1)^2} \bigg)du}$

By linearity :-

$\sf{=\dfrac{1}{8}\int \dfrac{1}{u^2-2u-1}du-\dfrac{1}{8}\int\dfrac{1}{u^2+1}du-\dfrac{1}{4}\int\dfrac{u-1}{(u^2+1)^2}du}$

$\boxed{\sf{Solving\ \dfrac{1}{u^2-2u-1}du:-}}$

$\sf{=\int \dfrac{1}{(u-\sqrt{2}-1)(u+\sqrt{2}-1)}du}$

By partial fraction decomposition,

$\sf{=\int \bigg(\dfrac{1}{2^\frac{3}{2}(u-\sqrt{2}-1)}-\dfrac{1}{2^\frac{3}{2}(u+\sqrt{2}-1)} \bigg)du}$

By linearity,

$\sf{=\dfrac{1}{2^\frac{3}{2}}\int \dfrac{1}{u-\sqrt{2}-1 }du-\dfrac{1}{2^\frac{3}{2} }\int\dfrac{1}{u+\sqrt{2}+1 }du}$

$\boxed{\sf{Solving\ \int\dfrac{1}{u-\sqrt{2}-1 }du:-}}$

Substitute  v  =  u  - √2 - 1

$\sf{\dfrac{dv}{du}=1 \longrightarrow du=dv}$

$\sf{=\int \dfrac{1}{v}dv}$

This is a standard integral:

= ln ( v )

Undo substitution v = u - √2 - 1 :-

$\sf{=In(u-\sqrt{2}-1) }$

$\boxed{\sf{Solving\ \int\dfrac{1}{u+\sqrt{2}-1 }du:-}}$

Substitute v = u + √2 - 1.

$\sf{\longrightarrow \dfrac{dv}{du}=1}\\\\\sf{\longrightarrow du=dv}$

$\sf{=\int \dfrac{1}{v}dv}$

Use previous result:

= In(v)

Undo substitution v = u + √2 - 1 :-

$\sf{=ln(u+\sqrt{2}-1)}$

Putting solved integrals :-

$\sf{=\dfrac{1}{2^\frac{3}{2}}\int \dfrac{1}{u-\sqrt{2}-1 }du-\dfrac{1}{2^\frac{3}{2} }\int\dfrac{1}{u+\sqrt{2}+1 }du}$

$\sf{=\dfrac{In(u-\sqrt{2}-1)}{2^\frac{3}{2}}-\dfrac{ln(u+\sqrt{2}-1)}{2^\frac{3}{2}}}$

$\boxed{\sf{Now\ solving\ \dfrac{1}{u^2+1}du}:-}$

This is a standard integral:

$\sf{=arctan(u) \:\:Or\:\:tan^{-1}(u)}$

${\boxed{\sf{Now\ solving\ \int\dfrac{u-1}{(u^2+1)^2}du:-}}$

Expanding :-

$\sf{=\int \bigg(\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^2} \bigg)du}$

Applying linearity :-

$\sf{= \int \dfrac{u}{(u^2+2)^2}du-\int\dfrac{1}{(u^2+1)^2}du}$

$\rule{150}{1}$

Kindly check the attachment for the next steps :-

$\underline{\rule{200}{2}}$

• *In integral calculus, the Weierstrass substitution or tangent half-angle substitution is a method for evaluating integrals, which converts a rational function of trigonometric functions of θ into an ordinary rational function of t by setting t = tan (θ/2).

• ln (Natural logarithm) :- The natural logarithm of a number is its logarithm to the base of the mathematical constant e, where e is an irrational and transcendental number approximately equal to 2.718281828459. The natural logarithm of x is generally written as ln x, loge x, or sometimes, if the base e is implicit, simply log x.

a49d80cb99c6ab3d620bbd1a72adcdd8.pdf

Answer 2

To Find:

Value of

$\rm{\displaystyle\int\limits^{\frac{\pi}{2}}_0{\frac{sin^{2}x}{sinx.cosx}}\,dx}$

Solution:

We know that,

$\longrightarrow\displaystyle\int\rm{{tanx}\,dx=ln(secx)+c}$

$\rule{190}{1}$

Let

$I=\displaystyle\int\limits^{\frac{\pi}{2}}_0\rm{{\frac{sin^{2}x}{sinx.cosx}}\,dx}$

$I=\displaystyle\int\limits^{\frac{\pi}{2}}_0\rm{{\frac{\cancel{sin^{2}x}}{\cancel{sinx}.cosx}}\,dx}$

$I=\displaystyle\int\limits^{\frac{\pi}{2}}_0\rm{{\frac{sinx}{cosx}}\,dx}$

$I=\displaystyle\int\limits^{\frac{\pi}{2}}_0\rm{{tanx}\,dx}$

$\rm{I=\Big[ln(secx)\Big]^{\frac{\pi}{2}}_{0}}$

$\rm{I=ln\bigg(sec\dfrac{\pi}{2}\bigg)-ln(sec\:0)}$

$\rm{I=ln(\infty)-ln(1)}$

$\rm{I=ln(\infty)-0}$

$\rm{I=ln(\infty)}$

Since, the above expression tends to infinity

So, integration of given function can't be determined in the

given interval.