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use standard identities first and then reduc this to an easy expression eqn. then integearte it. Use, tan(A+B) = (tanA + tanB)/(1 – tanAtanB) tanAtanB = 1 – (tanA + tanB)/tan(A+B) for tan2x and tanx, ...
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Detailed solution below------------------------------------------------------------------------------------- \quad \frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } { tan }^{ 2 }\frac { x }{ 2 } +1 \right)...
solution below.................................................................................................
Please find the answer, this could be easily solved if you know differentaition of inverse trignometric functions
tan(a) = (tanx – tan(x – a))/(1 + tanx*tan(x-a)) tanx/tana – tan(x – a)/tana – 1 = tanx*tan(x-a) Now see u can easily integerate L.HS. as tana remains as aconstant. Use by part wih tanx and tan(x-a)...
DEAR NAG, A VECTOR SPACE IS THE SET OF ANY NUMBERS THAT SATISFIES THE TEN PROPERTIES OF VECTOR ADDITION AND SCALR MULTIPLICATION.
First you have ask yourself that are you able to solve all problems of NCERT including excmpler problems without any help. If yes you are already prepared for JEE MAINS. If no then you will have to...
sorry made a mistake currect answer is 1/sinx =cosec x integration of cosec(x) is ln|cosec x-cot x|+c where c is arbitari constant....
mathamatical defination ..the computation of a definite integral, a fundamental concept of calculus, which allows, among many other uses, computing areas and averaging continuous functions. genaral...
they give sin(x)=x/2 0 is the one solution because sin(0)=0 so one solution exist for this question if you take pi/2 or pi/4 you have to p[rove that sin pi/2=pi/4 or sin pi/8=pi/16 which is greater...
one unit along y-axis in the negative direction. Thus, for each positive value of C, each parabola of the family has its vertex on the positive side of the y-axis and for negative values of C, each...
xy-3x-2y-10=0Take y common,y(x – 2) = 3x + 10y = (3x + 10)/(x-2)Point of discontinuity is at x = 2It represents a curve of which relation. in y and x with other than the ellipse, parabola, hyperbola...
integtration of =(x^2 – 2x +3) / x^4 =(x^-2-2x^-3+3x^-4) =integration of x^-2 dx-integration of -2x^-3 dx+integration of 3x^-4 =(x^-2+1/-2+1)-(2x^-3+1/-3+1)+(3x^-4+1/-4+1) =-1/x+1/x^2-1/x^3+c where c ...
the answer is indefinite integral of f(x) = integral of(x^2 + 3)/((x^6)*(x^2 + 1)) -3/5x^5 + 2/3x^3 – 2/x – 2arctanx + c where c is an arbitrary constant
apply h-allopital two times. lt x-->0 ( a x -b x ) => lt x-->0 ((loga)^2 a x -(logb)^2b x )/2 => (loga)^2 – (logb)^2 x 2i
MAXIMA MINIMA are calcultaed by differentiating the eqns. and equating it to zero. and we then calcuate the soln.s .i.e. values at which the slope is zero. We again check double differentiation, if...
let sinx = t^2 cosxdx = 2tdt. dx = 2tdt/root(1 – t^4) substitute it then integerate it. it becomes as: 2*(t^2(1-t^4)^-1/2)dt.
∫sinxcosxcos2xcos4xcos8xcos16x dx= > 1/2(∫sin2xcos2xcos4xcos8xcos16x )dx => (1/4)∫sin4xcos4xcos8xcos16x dx => (1/8) ∫sin8xcos8xcos16x dx => (1/16) ∫sin16xcos16x dx => (1/32) ∫sin32x dx =>...
dI = xroot(x-3)dx + 2root(x-3)dx Integerating both sides, integerate 2 nd term as usual and apply now by – part in first term. .i.e. in xroot(x-3) simply.
1/((x+1)^2*(x-1)^4)^1/3 => Integration is: (1/144)*( – 4(3x^3 – 6x^2 + x + 4)/(x-1)^3(x+1) – 6log(x-1) + 6log(x+1))
Write I = dx/x.(x^2n-a^2n)^1/2 = dx/x.(x^n + a^n)^1/2.(x^n – a^n)^1/2 = A/x + B/(x^n + a^n) + C/(x^n - a^n) Solve it by partially no. n is actually pi in image check it again plz.
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