Download, print and study this document offline |
Page 1 F. INTEGRATION BY REDUCTION FORMULAE Ex.47 If I n = ? ? 2 2 n x a x dx, prove that I n = – ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( ) 1 n ( ? ? a 2 I n–2 . Sol. I n = ? ? 2 2 n x a x dx = ? ? ? dx } x a x .{ x 2 2 1 n Applying integration by parts we get = x n–1 . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? + n 2 (n 1)x ? ? ? . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? dx = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 ) 1 n ( ? ? ? ? ) x a .( x 2 2 2 n 2 2 x a ? dx ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 – 3 ) 1 n ( ? I n ? I n + 3 ) 1 n ( ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? ? ? ? ? ? ? ? 3 2 n I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? I n = ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( a ) 1 n ( 2 ? ? I n–2 Page 2 F. INTEGRATION BY REDUCTION FORMULAE Ex.47 If I n = ? ? 2 2 n x a x dx, prove that I n = – ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( ) 1 n ( ? ? a 2 I n–2 . Sol. I n = ? ? 2 2 n x a x dx = ? ? ? dx } x a x .{ x 2 2 1 n Applying integration by parts we get = x n–1 . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? + n 2 (n 1)x ? ? ? . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? dx = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 ) 1 n ( ? ? ? ? ) x a .( x 2 2 2 n 2 2 x a ? dx ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 – 3 ) 1 n ( ? I n ? I n + 3 ) 1 n ( ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? ? ? ? ? ? ? ? 3 2 n I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? I n = ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( a ) 1 n ( 2 ? ? I n–2 INDEFINITE INTEGRATION Ex.48 Integration of 1/(x 2 + k) n . Sol. Thus ? ? ? 1 n 2 ) k x ( 1 1 . dx 1 n 2 ) k x ( x ? ? – 2 n (n 1) x. (x k) ? ? ? ? . 2x dx or I n–1 = 1 n 2 ) k x ( x ? ? + 2(n – 1) ? ? ? ? n 3 2 ) k x ( k ) k x ( dx, [ ? x 2 = (x 2 + k) – k] or I n–1 = 1 n 2 ) k x ( x ? ? + 2 (n – 1) ? ? ? ? ? ? ? ? ? ? ? ? ? ? n 2 1 n 2 ) k x ( dx k ) k x ( dx or I n–1 = 1 n 2 ) k x ( x ? ? + 2 (n – 1) l n–1 – 2k (n –1) l n . ? ? ? ? ? 2k (n–1) l n = 1 n 2 ) k x ( x ? ? + {2(n–1) – 1} l n–1 or 2k(n – 1) l n = 1 n 2 ) k x ( x ? ? + (2n – 3) I n–1 . Hence ? ? ? 1 n 2 ) k x ( dx = 1 n 2 ) k x )( 1 n ( k 2 x ? ? ? + ) 1 n ( k 2 ) 3 n 2 ( ? ? ? ? ? 1 n 2 ) k x ( dx . Above is the reduction formula for ? ? dx ] ) k x /( 1 [ n 2 . By repeated application of this formula the integral shall reduce to that of ) k x ( 1 2 ? which is k 1 tan –1 ? ? ? ? ? ? ? ? k x . Ex.49 Integrate (x + 2) / (2x 2 + 4x + 3) 2 . Sol. Here (d/dx) (2x 2 + 4x + 3) = 4x + 4. ? ? ? ? ? 2 2 ) 3 x 4 x 2 ( dx ) 2 x ( = ? ? ? ? ? ? 2 2 ) 3 x 4 x ( 2 1 2 ) 4 x 4 ( 4 1 dx = 4 1 ? ? ? ? 2 2 ) 3 x 4 x 2 ( dx ) 4 4 ( = ? ? ? ? ? ? ? ? ? ? 2 2 2 3 x 2 x dx ) 1 2 ( = 4 1 ? ? ? ? 2 2 ) 3 x 4 x 2 ( (4x + 4) dx + 4 1 ? ? ? ? ? ? ? ? ? 2 2 2 3 x 2 x dx = – ) 3 x 4 x 2 ( 4 1 2 ? ? + 4 1 ? ? ? ? ? ? ? ? ? 2 2 2 1 ) 1 x ( dx Now put x + 1 = t and then applying the reduction formula, we get I = ) 3 x 4 x 2 ( 4 1 2 ? ? + 4 1 c } ) 1 x ( 2 { tan 2 2 1 ) 1 x ( ) 1 x ( 1 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Ex.50 Integrate (2x + 3)/(x 2 + 2x + 3) 2 . Sol. Here (d/dx) (x 2 + 2x + 3) = 2x + 2 ? I = ? ? ? ? 2 2 ) 3 x 2 x ( ) 3 x 2 ( = 2 2 (2x 2 1)dx (x 2x 3) ? ? ? ? ? = ? ? ? ? 2 2 ) 3 x 2 x ( dx ) 2 x 2 ( = ? ? ? 2 2 ) 3 x 2 x ( dx = – ) 3 x 2 x ( 1 2 ? ? + ? ? ? 2 2 ) 3 x 2 x ( dx ....(i) Now let I 1 = ? ? ? 2 2 2 ] 2 ) 1 x [( dx (Put x + 1 = 2 tan t, so that dx = 2 sec 2 t dt) Page 3 F. INTEGRATION BY REDUCTION FORMULAE Ex.47 If I n = ? ? 2 2 n x a x dx, prove that I n = – ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( ) 1 n ( ? ? a 2 I n–2 . Sol. I n = ? ? 2 2 n x a x dx = ? ? ? dx } x a x .{ x 2 2 1 n Applying integration by parts we get = x n–1 . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? + n 2 (n 1)x ? ? ? . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? dx = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 ) 1 n ( ? ? ? ? ) x a .( x 2 2 2 n 2 2 x a ? dx ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 – 3 ) 1 n ( ? I n ? I n + 3 ) 1 n ( ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? ? ? ? ? ? ? ? 3 2 n I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? I n = ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( a ) 1 n ( 2 ? ? I n–2 INDEFINITE INTEGRATION Ex.48 Integration of 1/(x 2 + k) n . Sol. Thus ? ? ? 1 n 2 ) k x ( 1 1 . dx 1 n 2 ) k x ( x ? ? – 2 n (n 1) x. (x k) ? ? ? ? . 2x dx or I n–1 = 1 n 2 ) k x ( x ? ? + 2(n – 1) ? ? ? ? n 3 2 ) k x ( k ) k x ( dx, [ ? x 2 = (x 2 + k) – k] or I n–1 = 1 n 2 ) k x ( x ? ? + 2 (n – 1) ? ? ? ? ? ? ? ? ? ? ? ? ? ? n 2 1 n 2 ) k x ( dx k ) k x ( dx or I n–1 = 1 n 2 ) k x ( x ? ? + 2 (n – 1) l n–1 – 2k (n –1) l n . ? ? ? ? ? 2k (n–1) l n = 1 n 2 ) k x ( x ? ? + {2(n–1) – 1} l n–1 or 2k(n – 1) l n = 1 n 2 ) k x ( x ? ? + (2n – 3) I n–1 . Hence ? ? ? 1 n 2 ) k x ( dx = 1 n 2 ) k x )( 1 n ( k 2 x ? ? ? + ) 1 n ( k 2 ) 3 n 2 ( ? ? ? ? ? 1 n 2 ) k x ( dx . Above is the reduction formula for ? ? dx ] ) k x /( 1 [ n 2 . By repeated application of this formula the integral shall reduce to that of ) k x ( 1 2 ? which is k 1 tan –1 ? ? ? ? ? ? ? ? k x . Ex.49 Integrate (x + 2) / (2x 2 + 4x + 3) 2 . Sol. Here (d/dx) (2x 2 + 4x + 3) = 4x + 4. ? ? ? ? ? 2 2 ) 3 x 4 x 2 ( dx ) 2 x ( = ? ? ? ? ? ? 2 2 ) 3 x 4 x ( 2 1 2 ) 4 x 4 ( 4 1 dx = 4 1 ? ? ? ? 2 2 ) 3 x 4 x 2 ( dx ) 4 4 ( = ? ? ? ? ? ? ? ? ? ? 2 2 2 3 x 2 x dx ) 1 2 ( = 4 1 ? ? ? ? 2 2 ) 3 x 4 x 2 ( (4x + 4) dx + 4 1 ? ? ? ? ? ? ? ? ? 2 2 2 3 x 2 x dx = – ) 3 x 4 x 2 ( 4 1 2 ? ? + 4 1 ? ? ? ? ? ? ? ? ? 2 2 2 1 ) 1 x ( dx Now put x + 1 = t and then applying the reduction formula, we get I = ) 3 x 4 x 2 ( 4 1 2 ? ? + 4 1 c } ) 1 x ( 2 { tan 2 2 1 ) 1 x ( ) 1 x ( 1 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Ex.50 Integrate (2x + 3)/(x 2 + 2x + 3) 2 . Sol. Here (d/dx) (x 2 + 2x + 3) = 2x + 2 ? I = ? ? ? ? 2 2 ) 3 x 2 x ( ) 3 x 2 ( = 2 2 (2x 2 1)dx (x 2x 3) ? ? ? ? ? = ? ? ? ? 2 2 ) 3 x 2 x ( dx ) 2 x 2 ( = ? ? ? 2 2 ) 3 x 2 x ( dx = – ) 3 x 2 x ( 1 2 ? ? + ? ? ? 2 2 ) 3 x 2 x ( dx ....(i) Now let I 1 = ? ? ? 2 2 2 ] 2 ) 1 x [( dx (Put x + 1 = 2 tan t, so that dx = 2 sec 2 t dt) INDEFINITE INTEGRATION ? I 1 = ? ? 2 2 2 ) 2 t tan 2 ( dt t sec 2 = 4 2 ? dt t cos 2 = 4 2 ? 2 1 (1 + cos 2t) dt = 8 2 [+ 2 1 sin 2t] + 8 2 [t + sin t cos t] + c Now tan t = 2 1 x ? . Therefore sin t = } 2 ) 1 x {( 1 x 2 ? ? ? = ) 3 x 2 x ( 1 x 2 ? ? ? , and cos t = ) 3 x 2 x ( 2 2 ? ? Also t = tan –1 ? ? ? ? ? ? ? ? ? 2 1 x . Hence I 1 = 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + 8 2 . )} 3 x 2 x {( 1 x 2 ? ? ? . ) 3 x 2 x ( 2 2 ? ? . = 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + 4 1 ) 3 x 2 x ( 1 x 2 ? ? ? ? I = – 3 x 2 x 1 2 ? ? + 4 1 3 x 2 x 1 x 2 ? ? ? + 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + c, from (i) = ) 3 x 2 x ( 4 4 1 x 2 ? ? ? ? + 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + c = ) 3 x 2 x ( 4 3 x 2 ? ? ? + 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + c Ex.51 If I m = ? ? m ) x cos x (sin dx, then show that mI m = (sinx + cosx) m–1 . (sinx – cosx) + 2 (m – 1) I m–2 Sol. ? I m = ? ? m ) x cos x (sin dx = ? ? ? 1 m ) x cos x (sin . (sinx + cosx) dx, applying integration by parts. = (sin x + cosx) m–1 (cosx + sin x) – m 2 (m 1)(sinx cosx) dx ? ? ? ? . (cosx – sin x) . (sinx – cosx) dx = (sinx + cosx) m–1 (sinx – cos x) + (m – 1) ? ? ? 2 m ) x cos x (sin (sinx + cos x) 2 . dx As we know, (sinx + cosx) 2 + (sinx – cosx) 2 = 2, ? I m = (sinx + cosx) m–1 (sinx – cosx) + (m – 1) ? ? ? 2 m ) x cos x (sin {2 – (sinx + cosx) 2 } dx = (sinx + cosx) m–1 (sin x – cos x) + (m – 1) ? ? ? 2 m ) x cos x (sin 2 dx – (m – 1) ? ? dx ) x cos x (sin m I m = (sinx + cosx) m–1 (sinx – cosx) + (m – 1)I m–2 – (m – 1)I m or (m – 1) I m + I m = (sinx + cosx) m–1 (sinx – cosx) + 2 (m – 1) I m–2 or mI m = (sinx + cosx) m–1 (sinx – cosx) + 2 (m – 1) I m–2 Ex.52 If I m,n = ? , dx . nx cos . x cos m show that (m + n) I m,n = cos m x. sin nx + m I (m–1, n–1) Sol. We have, I m,n = m cos x.cosnx ? dx = (cos m x) ? ? ? ? ? ? n nx sin ? ?1 m cos m (–sinx). n nx sin dx = n 1 cos m x . sinnx + n m m 1 cos x(sinx.sinnx)dx ? ? As we have cos (n – 1) x = cos nx cos x + sin nx . sinx Page 4 F. INTEGRATION BY REDUCTION FORMULAE Ex.47 If I n = ? ? 2 2 n x a x dx, prove that I n = – ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( ) 1 n ( ? ? a 2 I n–2 . Sol. I n = ? ? 2 2 n x a x dx = ? ? ? dx } x a x .{ x 2 2 1 n Applying integration by parts we get = x n–1 . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? + n 2 (n 1)x ? ? ? . 2 2 3/2 (a x ) 3 ? ? ? ? ? ? ? ? ? ? ? ? dx = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 ) 1 n ( ? ? ? ? ) x a .( x 2 2 2 n 2 2 x a ? dx ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 – 3 ) 1 n ( ? I n ? I n + 3 ) 1 n ( ? I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? ? ? ? ? ? ? ? 3 2 n I n = – 3 ) x a ( x 2 / 3 2 2 1 n ? ? + 3 a ) 1 n ( 2 ? I n–2 ? I n = ) 2 n ( ) x a ( x 2 / 3 2 2 1 n ? ? ? + ) 2 n ( a ) 1 n ( 2 ? ? I n–2 INDEFINITE INTEGRATION Ex.48 Integration of 1/(x 2 + k) n . Sol. Thus ? ? ? 1 n 2 ) k x ( 1 1 . dx 1 n 2 ) k x ( x ? ? – 2 n (n 1) x. (x k) ? ? ? ? . 2x dx or I n–1 = 1 n 2 ) k x ( x ? ? + 2(n – 1) ? ? ? ? n 3 2 ) k x ( k ) k x ( dx, [ ? x 2 = (x 2 + k) – k] or I n–1 = 1 n 2 ) k x ( x ? ? + 2 (n – 1) ? ? ? ? ? ? ? ? ? ? ? ? ? ? n 2 1 n 2 ) k x ( dx k ) k x ( dx or I n–1 = 1 n 2 ) k x ( x ? ? + 2 (n – 1) l n–1 – 2k (n –1) l n . ? ? ? ? ? 2k (n–1) l n = 1 n 2 ) k x ( x ? ? + {2(n–1) – 1} l n–1 or 2k(n – 1) l n = 1 n 2 ) k x ( x ? ? + (2n – 3) I n–1 . Hence ? ? ? 1 n 2 ) k x ( dx = 1 n 2 ) k x )( 1 n ( k 2 x ? ? ? + ) 1 n ( k 2 ) 3 n 2 ( ? ? ? ? ? 1 n 2 ) k x ( dx . Above is the reduction formula for ? ? dx ] ) k x /( 1 [ n 2 . By repeated application of this formula the integral shall reduce to that of ) k x ( 1 2 ? which is k 1 tan –1 ? ? ? ? ? ? ? ? k x . Ex.49 Integrate (x + 2) / (2x 2 + 4x + 3) 2 . Sol. Here (d/dx) (2x 2 + 4x + 3) = 4x + 4. ? ? ? ? ? 2 2 ) 3 x 4 x 2 ( dx ) 2 x ( = ? ? ? ? ? ? 2 2 ) 3 x 4 x ( 2 1 2 ) 4 x 4 ( 4 1 dx = 4 1 ? ? ? ? 2 2 ) 3 x 4 x 2 ( dx ) 4 4 ( = ? ? ? ? ? ? ? ? ? ? 2 2 2 3 x 2 x dx ) 1 2 ( = 4 1 ? ? ? ? 2 2 ) 3 x 4 x 2 ( (4x + 4) dx + 4 1 ? ? ? ? ? ? ? ? ? 2 2 2 3 x 2 x dx = – ) 3 x 4 x 2 ( 4 1 2 ? ? + 4 1 ? ? ? ? ? ? ? ? ? 2 2 2 1 ) 1 x ( dx Now put x + 1 = t and then applying the reduction formula, we get I = ) 3 x 4 x 2 ( 4 1 2 ? ? + 4 1 c } ) 1 x ( 2 { tan 2 2 1 ) 1 x ( ) 1 x ( 1 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Ex.50 Integrate (2x + 3)/(x 2 + 2x + 3) 2 . Sol. Here (d/dx) (x 2 + 2x + 3) = 2x + 2 ? I = ? ? ? ? 2 2 ) 3 x 2 x ( ) 3 x 2 ( = 2 2 (2x 2 1)dx (x 2x 3) ? ? ? ? ? = ? ? ? ? 2 2 ) 3 x 2 x ( dx ) 2 x 2 ( = ? ? ? 2 2 ) 3 x 2 x ( dx = – ) 3 x 2 x ( 1 2 ? ? + ? ? ? 2 2 ) 3 x 2 x ( dx ....(i) Now let I 1 = ? ? ? 2 2 2 ] 2 ) 1 x [( dx (Put x + 1 = 2 tan t, so that dx = 2 sec 2 t dt) INDEFINITE INTEGRATION ? I 1 = ? ? 2 2 2 ) 2 t tan 2 ( dt t sec 2 = 4 2 ? dt t cos 2 = 4 2 ? 2 1 (1 + cos 2t) dt = 8 2 [+ 2 1 sin 2t] + 8 2 [t + sin t cos t] + c Now tan t = 2 1 x ? . Therefore sin t = } 2 ) 1 x {( 1 x 2 ? ? ? = ) 3 x 2 x ( 1 x 2 ? ? ? , and cos t = ) 3 x 2 x ( 2 2 ? ? Also t = tan –1 ? ? ? ? ? ? ? ? ? 2 1 x . Hence I 1 = 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + 8 2 . )} 3 x 2 x {( 1 x 2 ? ? ? . ) 3 x 2 x ( 2 2 ? ? . = 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + 4 1 ) 3 x 2 x ( 1 x 2 ? ? ? ? I = – 3 x 2 x 1 2 ? ? + 4 1 3 x 2 x 1 x 2 ? ? ? + 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + c, from (i) = ) 3 x 2 x ( 4 4 1 x 2 ? ? ? ? + 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + c = ) 3 x 2 x ( 4 3 x 2 ? ? ? + 8 2 tan –1 ? ? ? ? ? ? ? ? ? 2 1 x + c Ex.51 If I m = ? ? m ) x cos x (sin dx, then show that mI m = (sinx + cosx) m–1 . (sinx – cosx) + 2 (m – 1) I m–2 Sol. ? I m = ? ? m ) x cos x (sin dx = ? ? ? 1 m ) x cos x (sin . (sinx + cosx) dx, applying integration by parts. = (sin x + cosx) m–1 (cosx + sin x) – m 2 (m 1)(sinx cosx) dx ? ? ? ? . (cosx – sin x) . (sinx – cosx) dx = (sinx + cosx) m–1 (sinx – cos x) + (m – 1) ? ? ? 2 m ) x cos x (sin (sinx + cos x) 2 . dx As we know, (sinx + cosx) 2 + (sinx – cosx) 2 = 2, ? I m = (sinx + cosx) m–1 (sinx – cosx) + (m – 1) ? ? ? 2 m ) x cos x (sin {2 – (sinx + cosx) 2 } dx = (sinx + cosx) m–1 (sin x – cos x) + (m – 1) ? ? ? 2 m ) x cos x (sin 2 dx – (m – 1) ? ? dx ) x cos x (sin m I m = (sinx + cosx) m–1 (sinx – cosx) + (m – 1)I m–2 – (m – 1)I m or (m – 1) I m + I m = (sinx + cosx) m–1 (sinx – cosx) + 2 (m – 1) I m–2 or mI m = (sinx + cosx) m–1 (sinx – cosx) + 2 (m – 1) I m–2 Ex.52 If I m,n = ? , dx . nx cos . x cos m show that (m + n) I m,n = cos m x. sin nx + m I (m–1, n–1) Sol. We have, I m,n = m cos x.cosnx ? dx = (cos m x) ? ? ? ? ? ? n nx sin ? ?1 m cos m (–sinx). n nx sin dx = n 1 cos m x . sinnx + n m m 1 cos x(sinx.sinnx)dx ? ? As we have cos (n – 1) x = cos nx cos x + sin nx . sinx INDEFINITE INTEGRATION ? I m,n = n 1 cos m x . sin x + n m ? ? x cos 1 m . {cos(n–1) x – cosnx . cosx} dx = n 1 cos m x . sin x + n m ? ? ? dx . x ) 1 n cos( . x cos 1 m – n m ? nxdx cos . x cos m = n 1 cos m x . sin nx + n m I m–1,n–1 – n m I m,n ? I m,n + n m I m,n = n 1 [cos m x . sinnx + mI m–1, n–1 ] ? ? m n n ? ? ? ? ? ? ? I m,n = n 1 [cos m x . sin nx + mI m–1,n–1 ] ? (m + n) I m,n = cos m x . sin nx + mI m–1,n–1 .Read More
209 videos|443 docs|143 tests
|
209 videos|443 docs|143 tests
|
|
Explore Courses for JEE exam
|