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$\int_0^{\frac{\pi}{2}} \sin ^4 \theta \cos ^3 \theta d \theta=$
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$\frac{2}{35}$
We have $\int_0^{\frac{\pi}{2}}=\sin ^4 \theta \cos ^3 \theta \mathrm{d} \theta$
$\begin{aligned} & \text { Let } \mathrm{I}=\int_0^{\frac{\pi}{2}}=\sin ^4 \theta \cos ^3 \theta \mathrm{d} \theta \\ & \text { put since }=\mathrm{t} \Rightarrow \cos \theta \mathrm{d} \theta=\mathrm{dt} \\ & \text { at } \theta=0, \mathrm{t}=0 \\ & \theta=\frac{\neq}{2} \mathrm{t}=1 \\ & =\int_0^1 \mathrm{t}^2\left(1-\mathrm{t}^2\right) \mathrm{dt} \\ & \Rightarrow \int_0^1\left(\mathrm{t}^4-\mathrm{t}^6\right) \mathrm{dt} \\ & =\left[\frac{\mathrm{t}^5}{5}-\frac{\mathrm{t}^7}{7}\right]_0^1 \\ & =\frac{2}{35}\end{aligned}$
$\begin{aligned} & \text { Let } \mathrm{I}=\int_0^{\frac{\pi}{2}}=\sin ^4 \theta \cos ^3 \theta \mathrm{d} \theta \\ & \text { put since }=\mathrm{t} \Rightarrow \cos \theta \mathrm{d} \theta=\mathrm{dt} \\ & \text { at } \theta=0, \mathrm{t}=0 \\ & \theta=\frac{\neq}{2} \mathrm{t}=1 \\ & =\int_0^1 \mathrm{t}^2\left(1-\mathrm{t}^2\right) \mathrm{dt} \\ & \Rightarrow \int_0^1\left(\mathrm{t}^4-\mathrm{t}^6\right) \mathrm{dt} \\ & =\left[\frac{\mathrm{t}^5}{5}-\frac{\mathrm{t}^7}{7}\right]_0^1 \\ & =\frac{2}{35}\end{aligned}$
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