Question

求下列积分：（1）∫(cosx)^4 × (sinx)^3 dx （2）∫dx/(1+x^4)

Answer 1

(1) ∫(cosx)^4*(sinx)^3dx =-∫(cosx)^4*(sinx)^2dcosx =-∫(cosx)^4*[1-(cosx)^2]dcosx =∫(cosx)^6dcosx-∫(cosx)^4dcosx =(1/7)(cosx)^7-(1/5)(cosx)^5+C (2)∫[1/(1+x^4)]dx = 1/2∫[(x^2+1)-(x^2-1)]/(1+x^4)dx = 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/(1+x^4)dx } = 1/2 {∫(1+1/x^2)dx /(x^2+1/x^2) - ∫(1-1/x^2)dx/(x^2+1/x^2)} = 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -2] } = 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1} - ∫d(x+1/x) /[(x+1/x)^2 -2] } = 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1} - 1/2√2 ∫d[(x+1/x) /√2] [ 1/{[(x+1/x)/√2] -1} - 1/{[(x+1/x)/√2] +1 }] =√2/4*arctan[(x-1/x)/√2] - √2/8*ln|(x^2-x√2+1)/(x^2+x√2 +1)| + C