Question

Find the value of $\large \displaystyle \sf {\int \limits_0^{\pi} \bigg( \int \limits_{2\sin \theta} ^{4 \sin \theta} {r}^{3} \: dr\bigg) \: d \theta}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: \displaystyle \sf {\int \limits_0^{\pi} \bigg( \int \limits_{2\sin \theta} ^{4 \sin \theta} {r}^{3} \: dr\bigg) \: d \theta}$

We have to first integrate with respect to r, we get

$\rm \:  =  \: \displaystyle \sf\int \limits_0^{\pi}\bigg[\dfrac{ {r}^{3 + 1} }{3 + 1} \bigg] _{2\sin \theta} ^{4 \sin \theta} d\theta$

$\rm \:  =  \: \displaystyle \sf\int \limits_0^{\pi}\bigg[\dfrac{ {r}^{4} }{4} \bigg] _{2\sin \theta} ^{4 \sin \theta} d\theta$

$\rm \:  =  \: \dfrac{1}{4} \displaystyle \sf\int \limits_0^{\pi}(256 {sin}^{4}\theta - 16 {sin}^{4}\theta) d\theta$

$\rm \:  =  \: \dfrac{1}{4} \displaystyle \sf\int \limits_0^{\pi}240 \: {sin}^{4}\theta \: d\theta$

$\rm \:  =  \: 60 \displaystyle \sf\int \limits_0^{\pi} \: {sin}^{4}\theta \: d\theta$

We know,

$\boxed{\sf{ \int_0^{2a}f(x)dx = 2\int_0^{a}f(x)dx \: \: \: if \: f(2a - x) = f(x)}}$

So, using this can be rewritten as

$\rm \:  =  \: 60 \times 2\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{4}\theta \: d\theta$

$\rm \:  =  \: 120\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{4}\theta \: d\theta$

Using Walli's Formula, we get

$\rm \:  =  \: 120 \times \dfrac{3 \times 1}{4 \times 2} \times \dfrac{\pi}{2}$

$\rm \:  =  \: 15 \times 3 \times \dfrac{\pi}{2}$

$\rm \:  =  \: \dfrac{45\pi}{2}$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle \sf {\int \limits_0^{\pi} \bigg( \int \limits_{2\sin \theta} ^{4 \sin \theta} {r}^{3} \: dr\bigg) \: d \theta} = \frac{45\pi}{2} }}$

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Walli's Formula

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 2}{n(n - 2)(n - 4) - - - - 1} \: if \: n \: is \: odd$

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {cos}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 2}{n(n - 2)(n - 4) - - - - 1} \: if \: n \: is \: odd$

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - - - 2} \times \dfrac{\pi}{2} \: if \: n \: is \: even$

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {cos}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - - - 2} \times \dfrac{\pi}{2} \: if \: n \: is \: even$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: \displaystyle \sf {\int \limits_0^{\pi} \bigg( \int \limits_{2\sin \theta} ^{4 \sin \theta} {r}^{3} \: dr\bigg) \: d \theta}$

We have to first integrate with respect to r, we get

$\rm \:  =  \: \displaystyle \sf\int \limits_0^{\pi}\bigg[\dfrac{ {r}^{3 + 1} }{3 + 1} \bigg] _{2\sin \theta} ^{4 \sin \theta} d\theta$

$\rm \:  =  \: \displaystyle \sf\int \limits_0^{\pi}\bigg[\dfrac{ {r}^{4} }{4} \bigg] _{2\sin \theta} ^{4 \sin \theta} d\theta$

$\rm \:  =  \: \dfrac{1}{4} \displaystyle \sf\int \limits_0^{\pi}(256 {sin}^{4}\theta - 16 {sin}^{4}\theta) d\theta$

$\rm \:  =  \: \dfrac{1}{4} \displaystyle \sf\int \limits_0^{\pi}240 \: {sin}^{4}\theta \: d\theta$

$\rm \:  =  \: 60 \displaystyle \sf\int \limits_0^{\pi} \: {sin}^{4}\theta \: d\theta$

We know,

$\boxed{\sf{ \int_0^{2a}f(x)dx = 2\int_0^{a}f(x)dx \: \: \: if \: f(2a - x) = f(x)}}$

So, using this can be rewritten as

$\rm \:  =  \: 60 \times 2\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{4}\theta \: d\theta$

$\rm \:  =  \: 120\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{4}\theta \: d\theta$

Using Walli's Formula, we get

$\rm \:  =  \: 120 \times \dfrac{3 \times 1}{4 \times 2} \times \dfrac{\pi}{2}$

$\rm \:  =  \: 15 \times 3 \times \dfrac{\pi}{2}$

$\rm \:  =  \: \dfrac{45\pi}{2}$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle \sf {\int \limits_0^{\pi} \bigg( \int \limits_{2\sin \theta} ^{4 \sin \theta} {r}^{3} \: dr\bigg) \: d \theta} = \frac{45\pi}{2} }}$

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Walli's Formula

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 2}{n(n - 2)(n - 4) - - - - 1} \: if \: n \: is \: odd$

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {cos}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 2}{n(n - 2)(n - 4) - - - - 1} \: if \: n \: is \: odd$

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {sin}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - - - 2} \times \dfrac{\pi}{2} \: if \: n \: is \: even$

$\displaystyle \sf\int \limits_0^{\dfrac{\pi}{2} } \: {cos}^{n}\theta d\theta = \dfrac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - - - 2} \times \dfrac{\pi}{2} \: if \: n \: is \: even$