Question

Find the integration of
$\displaystyle \sf \: \int \frac{ \sin^{2}x - \cos ^{2} x}{ \sin \: x. \cos \: x}$

Lo karo answer mera level ka nhi hai. Google se mila​

Answer 1

$\LARGE{\bf{\underline{\underline\color{blue}{GIVEN:-}}}}$

$\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}$

$\LARGE{\bf{\underline{\underline\color{blue}{SOLUTION:-}}}}$

LHS:

$\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→$

Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.

$\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→$

Rearrange the terms.

$\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→$

We know that cos²A+sin²A=1.

$\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→$

Now here, take -2cos common from the numerator and +2cos common from the denominator.

$\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→$

Now, rearrange the terms, add 1 and 1 and take 2 common.

$\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→$

$\to\sf\dfrac{2+2sin-2cos}{sin+1}→$

Take 2 common.

$\to \sf \dfrac{ 2(1+sin) -2cos(sin+1) }{ 2(1+sin) + 2cos(sin +1 ) }→$

$\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→ </p><p>$

LHS=RHS.

HENCE PROVED!

FUNDAMENTAL TRIGONOMETRIC RATIOS:

$\begin{gathered}\begin{gathered}\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}\end{gathered}\end{gathered}$

T-RATIOS:

$\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}$

Answer 2

Answer:

★ Formula Applied :

$\begin{gathered}\sf \bullet\ \; cos^2x-sin^2x=cos2x\\\\\to\ \sf \pink{sin^2x-cos^2x=-cos2x}\end{gathered}$

$\begin{gathered}\bullet\ \; \sf sin2x=2.sinx.cosx\\\\\to \sf \blue{sinx.cosx=\dfrac{sin2x}{2}}\end{gathered}$

$\displaystyle \bullet\ \; \sf \int \dfrac{dx}{x}$

$\bullet\ \; \sf \ln (ab)=\ln a+\ln$

★ Explanation :

$\begin{gathered}\displaystyle \sf \int \dfrac{sin^2x-cos^2x}{sinx.cosx}dx\\\\\to \sf \int \dfrac{-cos2x}{\frac{sin2x}{2}}dx\\\\\to \sf \int \dfrac{-2cos2x}{sin2x}dx\end{gathered}$

Lets use substitution method ,

Let , u = sin2x

⇒ du = 2.cos2x.dx

$\begin{gathered}\to \displaystyle \sf \int \dfrac{-du}{u}\\\\\to \sf -\int \dfrac{du}{u}\\\\\to \sf -ln|u|\\\\\end{gathered}$

$\to \sf \red{-ln|sin2x|+c}$

$\to \sf - \ln |2sinx.cosx|+c$

$\to \sf - \ln |sinx.cosx|+(\ln 2+c)$

$\leadsto - \sf \red{\ln |sinx.cosx|+c}\ \; \bigstar$

★ Alternate Method :

$\displaystyle \sf \int \dfrac{sin^2x-cos^2x}{sinx.cosx}$

$\displaystyle \to \sf \int \left( \dfrac{sin^2x}{sinx.cosx}-\dfrac{cos^2x}{sinx.cosx}\right)dx$

$\begin{gathered}\displaystyle \to \sf \int \left( \dfrac{sinx}{cosx} -\dfrac{cosx}{sinx} \right)dx\\\\\to\ \sf \int (tanx-cotx)dx\\\\\to \sf \int tanx.dx-\int cotx.dx\\\\\to \sf ln|secx|-ln|sinx|+c\end{gathered}$

$\to \sf ln\left| \dfrac{secx}{sinx}\right|+c$

$\begin{gathered}\to \sf ln\left| \dfrac{1}{sinx.cosx} \right|+c\\\\\end{gathered}$]

$\leadsto \sf \pink{-ln|sinx.cosx|+c}\ \; \bigstar$