how to solve integration root(sin(x-a)/sin(x+a))dx
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This becomes, (sinxcosa - cosxsina / sinxcosa + cosxsina)1/2
ie (tanx - tana / tanx + tana)1/2
make the substitution tana + tanx = t2
So that the integral becomes 2(-2tana + t2)1/2/{1+{t2-tana)2}
Take t2 common form both the Nr and Dr, and make the substitution tana/t2 = y, so that 1/t3 dt = dy/(-2tana)
In the denominator you will have terms like, {y2/tan2a + (1-y)2}, which is a quadratic.
You will get a standard integral of the form ∫(√linear / Quadratic)
Which is integrated by the standard procedure.
Hope that helps.
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ie (tanx - tana / tanx + tana)1/2
make the substitution tana + tanx = t2
So that the integral becomes 2(-2tana + t2)1/2/{1+{t2-tana)2}
Take t2 common form both the Nr and Dr, and make the substitution tana/t2 = y, so that 1/t3 dt = dy/(-2tana)
In the denominator you will have terms like, {y2/tan2a + (1-y)2}, which is a quadratic.
You will get a standard integral of the form ∫(√linear / Quadratic)
Which is integrated by the standard procedure.
Hope that helps.
please mark brainliest to my answer when others answer if you liked it
also click thanks hope you understood and liked it :):):)
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