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By using the properties of definite integrals, evaluate the integrals`int_0^(2pi)cos^5x dx` |
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Answer» Let `I = int_0^(2pi) cos^5x dx` We know, `int_0^(2a) f(x) dx= 2 int_0^(a) f(x) dx` when `f(2a-x) dx = f(x)` Here, `cos^5(2pi-x) = cos^5x` `:. I = 2 int_0^pi cos^5x dx` Now, `int_0^(a) f(x) dx=0` when `f(a-x) dx = -f(x)` Here, `cos^5(pi-x) = - cos^5x` `:. I = 2 int_0^pi cos^5x dx = 2*0 = 0` `:. int_0^(2pi) cos^5x dx = 0.` |
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