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Grade 7Trigonometry

EVALUATE 1+cos2x/1-cos2x ? AND INTEGRATION OF SEC^3?

Profile image of satlapally pooja
10 Years agoGrade 7
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1 Answer

Profile image of jagdish singh singh
10 Years ago
\hspace{-1.2cm}$Evaluation of $\bf{\frac{1+\cos 2x}{1-\cos 2x} = \frac{2\cos^2 x}{2\sin^2 x} = \cot^2 x}$
 
\hspace{-0.6cm}$Let $\bf{I = \int \sec^3 xdx = \int \sec x\cdot \sec^2 xdx}$\\\\ Now using Integration by parts, We get\\\\ $\bf{I = \sec x\cdot \tan x-\int \sec x\tan x\cdot \tan xdx}$\\\\So $\bf{I = \sec x\tan x-\int \sec^3 xdx+\int \sec xdx}$\\\\So $\bf{I = \frac{1}{2}\sec x\tan x+ \frac{1}{2}\ln\left|\sec x+\tan x\right|+\mathcal{C}}$

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