Question

Evaluate the following:
(a) $\int\limits^\pi _0 {cos^{2}xsin^{4} x } \, dx$
(b)$\int\limits^4_0 \int\limits^√x_0 {\frac{1}{16} } \, dy dx$

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{I=\int^{\pi}_{0}cos^{2}(x)\,sin^{4}(x)\,dx}$

$\displaystyle\tt{\implies\,I=2\int^{\frac{\pi}{2}}_{0}cos^{2}(x)\,sin^{4}(x)\,dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)}$

$\displaystyle\tt{\implies\,I=2\int^{\frac{\pi}{2}}_{0}cos^{2}\left(\dfrac{\pi}{2}-x\right)\,sin^{4}\left(\dfrac{\pi}{2}-x\right)\,dx}$

$\displaystyle\tt{\implies\,I=2\int^{\frac{\pi}{2}}_{0}sin^{2}\left(x\right)\,cos^{4}\left(x\right)\,dx\,\,\,\,\,\,\,\,\,\,\,\,...(2)}$

Adding (1) and (2),

$\displaystyle\tt{\implies\,2I=2\int^{\frac{\pi}{2}}_{0}cos^{2}(x)\,sin^{4}(x)\,dx+2\int^{\frac{\pi}{2}}_{0}sin^{2}(x)\,cos^{4}(x)\,dx}$

$\displaystyle\tt{\implies\,I=\int^{\frac{\pi}{2}}_{0}cos^{2}(x)\,sin^{4}(x)\,dx+\int^{\frac{\pi}{2}}_{0}sin^{2}(x)\,cos^{4}(x)\,dx}$

$\displaystyle\tt{\implies\,I=\int^{\frac{\pi}{2}}_{0}\big\{cos^{2}(x)\,sin^{4}(x)+sin^{2}(x)\,cos^{4}(x)\big\}\,dx}$

$\displaystyle\tt{\implies\,I=\int^{\frac{\pi}{2}}_{0}cos^{2}(x)\,sin^{2}(x)\big\{sin^{2}(x)+cos^{2}(x)\big\}\,dx}$

$\displaystyle\tt{\implies\,I=\int^{\frac{\pi}{2}}_{0}cos^{2}(x)\,sin^{2}(x)\,dx}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{4}\int^{\frac{\pi}{2}}_{0}4\,cos^{2}(x)\,sin^{2}(x)\,dx}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{4}\int^{\frac{\pi}{2}}_{0}\,sin^{2}(2x)\,dx}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{4}\int^{\frac{\pi}{2}}_{0}\,\dfrac{1-cos(4x)}{2}\,dx}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{8}\int^{\frac{\pi}{2}}_{0}\,1-cos(4x)\,dx}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{8}\int^{\frac{\pi}{2}}_{0}\,dx-\dfrac{1}{8}\int^{\frac{\pi}{2}}_{0}cos(4x)\,dx}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{8}\big[x\big]^{\frac{\pi}{2}}_{0}-\dfrac{1}{8}\left[\dfrac{sin(4x)}{4}\right]^{\frac{\pi}{2}}_{0}}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{8}\left[\dfrac{\pi}{2}-0\right]-\dfrac{1}{32}\left[sin\left(4\cdot\dfrac{\pi}{2}\right)-0\right]}$

$\displaystyle\tt{\implies\,I=\dfrac{\pi}{16}-\dfrac{1}{32}\cdot\,sin\left(2\pi\right)}$

$\displaystyle\tt{\implies\,I=\dfrac{\pi}{16}}$