Math, asked by TrapNation, 1 year ago

Formulas of integration of Sin²x , cos²x and Tan²x ?? Pls its urgent

Answers

Answered by MaheswariS
7

\bf\int\,sin^2x\;dx

\text{We know that the identity}

\boxed{\bf\;cos2x=1-2\,sin^2x}

\implies\;sin^2x=\frac{1-cosx}{2}

\int\,sin^2x\;dx=\int\,\frac{1-cosx}{2}\;dx

\int\,sin^2x\;dx=\frac{1}{2}\int\,(1-cosx)\;dx

\int\,sin^2x\;dx=\frac{1}{2}(x-sinx)+c

\implies\boxed{\bf\int\,sin^2x\;dx=\frac{1}{2}(x-sinx)+c}

\bf\int\,cos^2x\;dx

\text{We know that the identity}

\boxed{\bf\;cos2x=2\,cos^2x-1}

\implies\;cos^2x=\frac{1+cosx}{2}

\int\,cos^2x\;dx=\int\,\frac{1+cosx}{2}\;dx

\int\,cos^2x\;dx=\frac{1}{2}\int\,(1+cosx)\;dx

\int\,cos^2x\;dx=\frac{1}{2}(x+sinx)+c

\implies\boxed{\bf\int\,cos^2x\;dx=\frac{1}{2}(x+sinx)+c}

\bf\int\,tan^2x\;dx

\text{We know that the identity}

\boxed{\bf\;tan2x=sec^2x-1}

\int\,tan^2x\;dx=\int\,(sec^2x-1)\;dx

\int\,tan^2x\;dx=tanx-x+c

\implies\boxed{\bf\int\,tan^2x\;dx=tanx-x+c}

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