2023TrialQ4Which of the following is an expression for ∫dx7−6x−x2?\int \frac{dx}{\sqrt{7-6x-x^2}}?∫7−6x
2023TrialQ7Which expression is equal to ∫3xlnx dx?\int 3\sqrt{x}\ln x\,dx?∫3xlnxdx? (A) 2xx(lnx−23)+c\displaystyle 2x\sqrt{x}\left(\ln x-\frac{2}{3}\right)+c2xx(lnx−3 (B) 2xx(lnx+23)+c\displaystyle 2x\sqrt{x}\left(\ln x+\frac{2}{3}\right)+c2xx(lnx+3 (C) 1x(32lnx−1)+c\displaystyle \frac{1}{\sqrt{x}}\left(\frac{3}{2}\ln x-1\right)+cx1 (D) 1x(32lnx+1)+c\displaystyle \frac{1}{\sqrt{x}}\left(\frac{3}{2}\ln x+1\right)+cx11 markMultiple choice
2023TrialQ12 (b) (i)Show that ∫0π/2dx1+cosx+sinx=ln2.\int_0^{\pi/2}\frac{dx}{1+\cos x+\sin x}=\ln 2.∫0π/21+cosx+sinxdx=ln2.3 marksShort answer
2023TrialQ12 (c) (iii)Let In=∫e−11(1+logex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn=∫e−11(1+logex)ndx and Jn=∫e−11(logex)(1+logex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn=∫e for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,…. Hence find the value of J3J_3J3 in simplest exact form.2 marksShort answer
2023TrialQ12 (c) (i)Let In=∫e−11(1+logex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn=∫e−11(1+logex)ndx and Jn=∫e−11(logex)(1+logex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn=∫e for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,…. Show that In=1−nIn−1I_n=1-nI_{n-1}In=1−nIn−1 for n=1,2,3,….2 marksShort answer
2023TrialQ12 (b) (ii)By making the substitution u=a−xu=a-xu=a−x, show that ∫0af(x) dx=∫0af(a−x) dx.\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx.∫0af(x)dx=∫0af(a−x)dx.1 markShort answer
2023TrialQ12 (b) (iii)Hence or otherwise, evaluate ∫0π/2x1+cosx+sinx dx.\int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx.∫0π/21+cosx+sinxxdx.2 marksShort answer
2023TrialQ12 (c) (ii)Let In=∫e−11(1+logex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn=∫e−11(1+logex)ndx and Jn=∫e−11(logex)(1+logex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn=∫e for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,…. Show that Jn=1−(n+2)InJ_n=1-(n+2)I_nJn=1−(n+2)In for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,….2 marksShort answer
2023TrialQ13 (c) (i)Find real numbers aaa, bbb and ccc such that 10(x+1)(x2+4)≡ax+1+bx+cx2+4.\frac{10}{(x+1)(x^2+4)}\equiv\frac{a}{x+1}+\frac{bx+c}{x^2+4}.(x+1)(x2+4)10≡x+1a3 marksShort answer
2023TrialQ13 (c) (ii)Hence find ∫10(x+1)(x2+4) dx.\int \frac{10}{(x+1)(x^2+4)}\,dx.∫(x+1)(x2+4)10dx.2 marksShort answer
2022HSCQ4Of the following expressions, which one need NOT contain a term involving a logarithm in its anti-derivative? A. x+2x2+4x+5\dfrac{x+2}{x^2+4x+5}x2+4x+5x+2 B. x+2x2−4x−5\dfrac{x+2}{x^2-4x-5}x2−4x−5x+2 C. x−1x3−x2+x−1\dfrac{x-1}{x^3-x^2+x-1}x3−x2+x−1x−1 D. x+1x3−x2+x−1\dfrac{x+1}{x^3-x^2+x-1}x3−x2+x−1x+11 markMultiple choice
2022HSCQ5If ∫axf(t) dt=g(x)\displaystyle\int_a^x f(t)\,dt=g(x)∫axf(t)dt=g(x), which of the following is a primitive of f(x)g(x)f(x)g(x)f(x)g(x)? A. 12[f(x)]2\frac12[f(x)]^221[f(x)]2 B. 12[f′(x)]2\frac12[f'(x)]^221[f′(x)]2 C. 12[g(x)]2\frac12[g(x)]^221[g(x)]2 D. 12[g′(x)]2\frac12[g'(x)]^221[g′(x)]21 markMultiple choice
2022TrialQ5∫16x−x2 dx=\int\frac{1}{\sqrt{6x-x^2}}\,dx=∫6x−x21dx= A. sin−1(x6)+C\sin^{-1}\left(\frac{x}{\sqrt6}\right)+Csin−1(6 B. sin−1(x6)+C\sin^{-1}\left(\frac{x}{6}\right)+Csin−1(6x)+C C. sin−1(x−36)+C\sin^{-1}\left(\frac{x-3}{\sqrt6}\right)+Csin−1(6 D. sin−1(x−33)+C\sin^{-1}\left(\frac{x-3}{3}\right)+Csin−1(3x−3)+C1 markMultiple choice
2022TrialQ6If f(x)f(x)f(x) is even then ∫−aaf(x) dx=\int_{-a}^{a}f(x)\,dx=∫−aaf(x)dx= A. 2∫0af(a−x) dx2\int_0^a f(a-x)\,dx2∫0af(a−x)dx B. 2∫0−af(a−x) dx2\int_0^{-a} f(a-x)\,dx2∫0−af(a−x)dx C. 2∫a0f(x) dx2\int_a^0 f(x)\,dx2∫a0f(x)dx D. All of the above.1 markMultiple choice
2022TrialQ7Which of the following definite integrals has a positive value? A. ∫−1/21/2x3(1+cosx) dx\displaystyle\int_{-1/2}^{1/2}x^3(1+\cos x)\,dx∫−1/21/2x3(1+cosx)dx B. ∫−1/21/2sin−1x1+x4 dx\displaystyle\int_{-1/2}^{1/2}\frac{\sin^{-1}x}{1+x^4}\,dx∫−1/21/21+x C. ∫−1/21/2cos−1xe−x2 dx\displaystyle\int_{-1/2}^{1/2}\frac{\cos^{-1}x}{e^{-x^2}}\,dx∫−1/21/2e− D. ∫−1/21/2tan−1xex+e−x dx\displaystyle\int_{-1/2}^{1/2}\frac{\tan^{-1}x}{e^x+e^{-x}}\,dx∫−1/21/2e1 markMultiple choice
2022TrialQ11 (e)Find ∫15+4x−x2 dx.\int\frac{1}{\sqrt{5+4x-x^2}}\,dx.∫5+4x−x21dx.2 marksShort answer
2022HSCQ11 (f)Using the substitution t=tanx2t=\tan\frac{x}{2}t=tan2x, find ∫dx1+cosx−sinx.\int\frac{dx}{1+\cos x-\sin x}.∫1+cosx−sinxdx.3 marksShort answer
2022HSCQ11 (b)Evaluate ∫sin3(2x)cos(2x) dx\displaystyle\int \sin^3(2x)\cos(2x)\,dx∫sin3(2x)cos(2x)dx.2 marksShort answer
2022TrialQ12 (a) (ii)Hence, or otherwise, find ∫x2−2xx2−2x−15 dx.\int\frac{x^2-2x}{x^2-2x-15}\,dx.∫x2−2x−15x2−2xdx.2 marksShort answer
2022TrialQ12 (c) (ii)Hence find ∫−7x−18(x−1)(x2+4) dx.\int\frac{-7x-18}{(x-1)(x^2+4)}\,dx.∫(x−1)(x2+4)−7x−18dx.2 marksShort answer
2022TrialQ12 (b)Evaluate ∫0π/4sec4xtan2x dx.\int_0^{\pi/4}\sec^4x\tan^2x\,dx.∫0π/4sec4xtan2xdx.3 marksShort answer
2022TrialQ12 (e)Use the substitution x=2sinθx=2\sin\thetax=2sinθ to find ∫dx(4−x2)3/2.\int\frac{dx}{(4-x^2)^{3/2}}.∫(4−x2)3/2dx.3 marksShort answer
2022TrialQ12 (b)Find ∫12+cosx dx\displaystyle\int\frac{1}{2+\cos x}\,dx∫2+cosx1dx.3 marksShort answer