Changing the integral activity



NESA exemplar question solutionsIntegrating using substitutionExamples of integrals to be determined using given substitutions include:Use the substitution u=x2+1 to determine x1+x2dx Substitution:u=x2+1dudx=2xdu=2x?dxdu2=x?dxLet I=x1+x2dx=1+x2?x?dx=u?du2 =12u12du=12?u3232+C=u323+C=x2+1323+C=x2+133+CUse the substitution u=1+t to evaluate 01t1+tdtSubstitution:u=1+tu-1=tdudt=1du=dtBounds: t=0, u=1t=1, u=2Let I=01t1+tdt=12u-1udu=12uu-1udu=12u12-u-12du=2u323-2u1212=22323-2212-21323-2112=423-22-23-2=43-223 or 4-223 Use the substitution u=x2 to determinex9+x4dxSubstitution:u=x2dudx=2xdu2=x?dxLet I=x9+x4dx=19+x22?x?dx=19+u2?du2=12132+u2?du=12?13tan-1u3+C=16tan-1x23+CIntegrating powers of trigonometric functionsExamples of integrals to be determined involving squares of the trigonometric functions sine and cosine, and those that can be found by a simple substitution, include:0π4sin22xdxUse the identity: sin2nx=12(1-cos2nx) where n=2, i.e. sin22x=12(1-cos4x)Let I=0π4sin22xdx=0π412(1-cos4x)dx=12x-14sin4x0π4=12π4-14sin4?π4-0-14sin4?0=12π4-14sinπ-0-14sin0=12π4-0-0-0=π80π4sin2xcosxdxSubstitution:u=sinxdudx=cosxdu=cosx?dxBounds: x=0, u=0x=π4, u=12=22Let I=0π4sin2xcosxdx=0π4sin2x?cosx?dx=022u2?du=u33022=2233-033=2283=243=2120π4sin2xcos3xdxSubstitution:u=sinxdudx=cosxdu=cosx?dxBounds: x=0, u=0x=π4, u=12=22Let I=0π4sin2xcos3xdx=0π4sin2x?cos2x?cosx?dx=0π4sin2x?(1-sin2x)?cosx?dx=0π4sin2x?(1-sin2x)?cosx?dx=022u2?1-u2du=022u2-u4du=u33-u55022=2233-2255-033-055=2283-42325=212-240=72120Evaluate 4π12π4sin24xdxUse the identity: sin2nx=12(1-cos2nx) where n=4, i.e. sin24x=12(1-cos8x)Let I=4π12π4sin24xdx=4π12π4121-cos8xdx=2π12π41-cos8xdx=2x-18sin8xπ12π4=2π4-18sin8?π4-π12-18sin8?π12=2π4-18sin2π-π12+18sin2π3=2π4-0-π12+18?32=2π6+316=π3+38If d2ydx2=2cos2x and when x=π2, dydx=0, y=0, then find y in terms of x.Use the identity: cos2nx=12(1+cos2nx) where n=1, i.e. cos2x=12(1+cos2x)Alternatively, consider the double angle result for cosine and rearrange.d2ydx2=2cos2x dydx=2cos2x dx=2?12(1+cos2x) dx=(1+cos2x) dx=x+12sin2x+CSubstitute x=π2, dydx=00=π2+12sin2?π2+C0=π2+0+CC=-π2dydx= x+12sin2x-π2y=x+12sin2x-π2 dx=x22-14cos2x-π2x+kSubstitute x=π2, y=00=π222-14cos2?π2-π2?π2+k 0=π28-14?-1-π24+k0=-π28+14+k k=π28-14y=x22-14cos2x-π2x+π28-14Integrating inverse trigonometric functionsPractice is needed in the use of the product rule, quotient rule, and chain rule in relation to the inverse trigonometric functions. Students will need practice on questions involving the inverse trigonometric functions and integration by substitution.Differentiate sin-1x+cos-1x, and hence show that sin-1x+cos-1x=π2.Let y=sin-1x+cos-1xdydx=11-x2+-11-x2dydx=0This implies that the function y is a constant. To investigate the value of this constant:Let sin-1x=θ∴sinθ=x=cos?π2-θIf cos?π2-θ=xThen cos-1x=π2-θcos-1x=π2-sin-1xcos-1x+sin-1x=π2 as required.If y=cos-1x+cos-1(-x), find dydx and show that y=π for all x in the domain.dydx=-11-x2+11-(-x)2dydx=-11-x2+11-x2dydx=0This implies that the function y is a constant. To investigate the value of this constant:Now, let cos-1x=θ∴cosθ=x∴cos?(π-θ)=-x∴cos-1-x=π-θ∴cos-1-x=π-cos-1x∴cos-1x+cos-1-x=π as required.01dx4-x2Let I=01dx4-x2=01dx22-x2=sin-1x201=sin-112-sin-10=π601dx3-x2Let I=01dx3-x2=01dx32-x2=sin-1x301=sin-113-sin-10=0.615… 012dx1+4x2Let I=012dx1+4x2=120122?dx1+2x2=12tan-12x012=12tan-12?12-tan-10=12tan-11=12?π4=π8Other questionsProve that ddxxsin-1x=sin-1x+x1-x2. Hence use the substitution u=1-x2 to show that 012sin-1xdx=π12+32-1.Use the product rule to differentiate xsin-1xi.e. dydx=u?dvdx+v?dudx, where:u=xdudx=1v=sin-1xdvdx=11-x2ddxxsin-1x=x?11-x2+sin-1x?1=sin-1x+x1-x2∴ddxxsin-1x=sin-1x+x1-x2 as required. Given the above result, ddxxsin-1x-x1-x2=sin-1xLet I=012sin-1xdx=012ddxxsin-1x-x1-x2?dx=012ddxxsin-1x?dx-012x1-x2?dxSubstitution:u=1-x2dudx=-2x-du2=x?dxBounds:x=0, u=1x=12, u=34=012ddxxsin-1x?dx-1341u?-du2=012ddxxsin-1x?dx+12134u-12?du=xsin-1x012+122u12134=12sin-112-0sin-10+1223412-2112=12?π6+123-2=π12+123-2=π12+32-1 as required. ................
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