Wednesday, July 17, 2019

IGNOU MTE01 SOLVED ASSIGNMENT 2019 4(a)

\int {{{\cos }^n}x = \sin x{{\cos }^{n - 1}}x + \left( {n - 1} \right){{\sin }^2}x\int {{{\cos }^{n - 2}}x} } dx
\int {{{\cos }^n}x = \sin x{{\cos }^{n - 1}}x + \left( {n - 1} \right)} \int {{{\cos }^{n - 2}}xdx} + \int {{{\cos }^n}xdx} - n\int {{{\cos }^n}xdx}
n\int {{{\cos }^n}x = \sin x{{\cos }^{n - 1}}x + \left( {n - 1} \right)} \int {{{\cos }^{n - 2}}xdx}
\int {{{\cos }^n}x = \frac{1}{n}\sin x{{\cos }^{n - 1}}x + \frac{{\left( {n - 1} \right)}}{n}} \int {{{\cos }^{n - 2}}xdx}


Now we evaluate

\int {{{\cos }^6}x}

Thus

\int {{{\cos }^6}x = \frac{1}{6}\sin x{{\cos }^5}x + \frac{5}{6}} \int {{{\cos }^4}xdx}
\int {{{\cos }^6}x = \frac{1}{6}\sin x{{\cos }^5}x + \frac{5}{6}} \left( {\frac{{\sin x{{\cos }^5}x}}{4} + \frac{3}{4}\int {{{\cos }^2}xdx} } \right)
\int {{{\cos }^6}x = \frac{1}{6}\sin x{{\cos }^5}x + \frac{5}{{24}}} \sin x{\cos ^3}x + \frac{{15}}{{48}}\sin x\cos x


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