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chwala
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- I am looking at this definite integral; i will go slow on it with the whole intention being to understand the steps and how to apply. I will attempt to use the reduction method and i also noted that this can be approached using Beta and Gamma relationship. Its Time to upscale my intellect.
$$\int_0^π \sin^ n x dx = \int_0^π \sin^ {n-1} x\sin x dx$$
Letting ##u=\sin^{n-1} x## and ##v^{'}= \sin x## then,
$$\int_0^π \sin^ {n-1} x\sin x dx = [-\sin^ {n-1} x⋅\cos x]_0^π+ \int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x
dx$$
$$=\int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x dx $$
$$=\int_0^π (n-1)\sin^{n-2} x ⋅ (1- \sin^2 x) dx $$
Letting
$$I_n = \int_0^π \sin^ n x dx $$
$$I_n = (n-1)\int_0^π \left(\sin^{n-2} x - \sin^n x \right)dx $$
$$I_n = (n-1)(I_{n-2} - I_n)$$
$$nI_n = (n-1)I_{n-2}$$
Now up to this point it is quite clear. The next thing for me to check on is on conditions where ##n## is even or odd. Most importantly to understand the beta and gamma approach that forms the basis of this post.
Cheers Man!
Letting ##u=\sin^{n-1} x## and ##v^{'}= \sin x## then,
$$\int_0^π \sin^ {n-1} x\sin x dx = [-\sin^ {n-1} x⋅\cos x]_0^π+ \int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x
dx$$
$$=\int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x dx $$
$$=\int_0^π (n-1)\sin^{n-2} x ⋅ (1- \sin^2 x) dx $$
Letting
$$I_n = \int_0^π \sin^ n x dx $$
$$I_n = (n-1)\int_0^π \left(\sin^{n-2} x - \sin^n x \right)dx $$
$$I_n = (n-1)(I_{n-2} - I_n)$$
$$nI_n = (n-1)I_{n-2}$$
Now up to this point it is quite clear. The next thing for me to check on is on conditions where ##n## is even or odd. Most importantly to understand the beta and gamma approach that forms the basis of this post.
Cheers Man!
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