2cos' x sin xdx.计算例1Jo解令 t=cosx,dt = -sinxdx.元t=0,x=0=t=1x!-2儿25cosx sin xdx0
例1 计算 cos sin . 2 0 5 x xdx 解 令 t = cos x, 2 x = t = 0, x = 0 t = 1, 2 0 5 cos x sin xdx = − 0 1 5 t dt 1 0 6 6 t = . 6 1 = dt = −sin xdx
"sin' x - sin' xdx.例2计算解 f(x)= /sin x-sin'x = cosx(sinx)/sin' x - sin' xdx= (" Jcos x(sinx) dx=Je cos x(sinx)dx -" cos x(sinx)dx?Fe (sinx)d sinx -f"(sinx)id sinx2(sin x)(sinx)-=3泰元2
例2 计算 解 sin sin . 0 3 5 x − xdx f x x x 3 5 ( ) = sin − sin ( )2 3 = cos x sin x − 0 3 5 sin x sin xdx ( ) = 0 2 3 cos x sin x dx ( ) = 2 0 2 3 cos x sin x dx ( ) − 2 2 3 cos x sin x dx ( ) = 2 0 2 3 sin x d sin x ( ) − 2 2 3 sin x d sin x ( ) 2 0 2 5 sin 5 2 = x ( ) − 2 2 5 sin 5 2 x . 5 4 =
31dxCei例3计算J.e x/Inx(1-Inx)3d(lnx)24解原式：=ee Inx(1-Inx).33d(lnx)eyd/InxD=Jre JInx /(1-Inx)21.e1-(/lnx)=2[aresin(/Inx)], =".合
例3 计算 解 . ln (1 ln ) 4 3 − e e x x x dx 原式 − = 4 3 ln (1 ln ) e (ln ) e x x d x − = 4 3 ln (1 ln ) e (ln ) e x x d x − = 4 3 2 1 ( ln ) ln 2 e e x d x 4 3 2 arcsin( ln ) e e = x . 6 =
1Cadx.(a>0)例4计算Jo22-xx+va'解令x=asint,dx = acostdt.元x=0=t=0tx=a==2元acost原式=dtasint+a'(1-sin't)元costcost-sint2dtdt1+JOsint +costsint + cost-++ mino -合
例4 计算 解 + − a dx a x a x 0 2 2 . ( 0) 1 令 x = asint, x = a , 2 t = x = 0 t = 0, dx = acostdt, 原式 + − = 2 0 2 2 sin (1 sin ) cos dt a t a t a t + = 2 0 sin cos cos dt t t t + − = + 2 0 sin cos cos sin 1 2 1 dt t t t t 2 0 lnsin cos 2 1 2 2 1 + + = t t . 4 =
例5 当f(x)在-a,al上连续，且有①f(x)为偶函数，则" f(x)dx = 2f" f(x)dx ;②f(x)为奇函数，则["f(x)dx =0.证" f(x)dx = F", f(x)dx+ f" f(x)dx,在[", f(x)dx中令x=-t
例 5 当 f (x)在[−a, a]上连续，且有 ① f (x)为偶函数，则 − = a a a f x dx f x dx 0 ( ) 2 ( ) ； ② f (x)为奇函数，则− = a a f (x)dx 0. 证 ( ) ( ) ( ) , 0 0 − − = + a a a a f x dx f x dx f x dx 在− 0 ( ) a f x dx中令x = −t