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3 years ago

Marking as brainliest!! Agree or disagree with them

English

2 answers:

Natali5045456 [20]3 years ago

8 0

Answer:

Are you asking for our opinions? If so...

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BIG Agree

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Ket [755]3 years ago

6 0

Agree
agree
agree
agree
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I will make you brainlist if you help me the 2 question by explain them i swear i will make you brainlist

Xelga [282]

★ 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 :

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

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

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