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                                        11. NJEHSIMI INTEGRAL291USHTRIMEShembulli 8Të njehsohet ∫sin 2x xdx.ZgjidhjeVëmë re që dx = 12 d(2x). Prandaj ∫sin2 xdx = ∫sin2 x ·12 d(2x).Duke shënuar u = 2x, marrim ∫sin2 xdx = 12∫sinudu = 12 (–cos u) + c = – 12 · cos 2 x + cC Ushtrohuni duke zbatuarNjehsoni: a) ∫ (2x – 5)10 dx;b) ∫ (2 – 3x)7 dx;c) ∫ x(1 + x2)3 dx.1 Të njehsohen integralet:a) ∫ x4 – 3x2 – x – 1x3 dx; ∫x3 + 1x2 – x + 1 dx; ∫ x3 – 1x2 + x + 1 dx;b) ∫ e2x – 1ex – 1 dx; ∫ e2x – exex dx; ∫ 32x + 2 · 3xex dx; ∫a2x – 1ax + 1 dx;c) ∫(x2 – 1)2xdx; ∫(2 – x)3x2 dx; ∫(1 + x)3x3 dx;d) ∫cos x · tgxdx; ∫sinx · cot g xdx; ∫ cos x – sinxcos2 x + sin2 xdx; ∫ sin4 x – cos4 xsin x – cos xdx;e) ∫ 1x2 – 22 dx; ∫ 3x2 – 9 dx; ∫ 516 – x2 dx; ∫ 1x2 – 3 dx;2 Të gjenden integralet:a) ∫(x + 5)7 dx; ∫(2x – 3)4 dx; ∫(1 – 3x)5 dx; ∫(x + 2)32 dx.b) ∫ 5 – x dx; ∫ 3x – 2 3 dx; ∫ (4 – 7x) 5 3 dx;c) ∫ 1x – 2 dx ∫ 11 – 3x 3 dx ∫ 2(3 + 5x) 4 3 dxd) ∫ x(x2 + 2)3 dx; ∫ x(3x2 – 5)4 dx; ∫ x(3 – x2)23 dx.e) ∫ x2 x3 – 6 dx; ∫ x5 x6 + 9 6 dx; ∫(x + 1) (x + 1) 5 3 dx;f) ∫ xx2 – 5 dx ∫ x31 + x4 dx; ∫ x2(x3 + 1) 3 2 dx;

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