1.

integral cos^4x dx

Answer»

To find ,  \(\int \cos^4x dx\)

Since we know that 

\(\cos^2x = {1 \over 2}(\cos2x+1)\)

so , \(\cos^4x = {1 \over 4}(\cos^22x+1+2\cos2x)\)

We can also expand \(\cos^22x \)

⇒ \(\cos^22x = {1 \over 2}(\cos4x + 1)\)

Which means \(\cos^4x = {1 \over 8}(\cos4x) + {1 \over 2}(\cos2x) + {3 \over 8}\)

Now we have to evaluate the integral of : 

\({1 \over 8}\int\cos4xdx + {1 \over 2}\int\cos2xdx + {3 \over 8}\int dx\)

Since , Integral of \(\cos x = \sin x\)

⇒\({1 \over 8}\times{1 \over 4}\int\cos4x (d(4x)) + {1 \over 2}\times{1 \over 2}\int\cos2x(d(2x)) + {3 \over 8}\int dx\)

⇒\({\sin4x \over 32} + {\sin2x \over 4} + {3x \over 8} + C\)

Hence , \(\int \cos^4x dx\) = \({\sin4x \over 32} + {\sin2x \over 4} + {3x \over 8} + C\)


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