∫ ( e x ln ⁡ a − e a ln ⁡ x + 3 e a ln ⁡ a ) d x = \int_{ }^{ }\left(e^{x\ln a}-e^{a\ln x}+3e^{a\ln a}\right)dx= ∫ ( e x ln a − e a ln x + 3 e a ln a ) d x =

a x ln ⁡ a − x a + 1 a + 1 + 3 x + c \frac{a^x}{\ln a}-\frac{x^{a+1}}{a+1}+3x+c ln a a x − a + 1 x a + 1 + 3 x + c

a x ln ⁡ a + x a a + 1 + 3 + c \frac{a^x}{\ln a}+\frac{x^a}{a+1}+3+c ln a a x + a + 1 x a + 3 + c

a x ln ⁡ a + x a + 1 a + c \frac{a^x}{\ln a}+\frac{x^{a+1}}{a}+c ln a a x + a x a + 1 + c

If ∫ sin ⁡ x sin ⁡ ( x − α ) d x = \int_{ }^{ }\frac{\sin x}{\sin\left(x-\alpha\right)}dx\ = ∫ sin ( x − α ) sin x d x = A x + B log ⁡ sin ⁡ ( x − α ) + C Ax+B\log\sin\left(x-\alpha\right)+C\ A x + B lo g sin ( x − α ) + C t h e n ( A , B ) = then\ \left(A,B\right)= t h e n ( A , B ) =

( cos ⁡ α , sin ⁡ α ) \left(\cos\alpha,\sin\alpha\right) ( cos α , sin α )

( sin ⁡ α , cos ⁡ α ) \left(\sin\alpha,\cos\alpha\right) ( sin α , cos α )

( sin ⁡ α , − cos ⁡ α ) \left(\sin\alpha,-\cos\alpha\right) ( sin α , − cos α )

( − cos ⁡ α , − sin ⁡ α ) \left(-\cos\alpha,-\sin\alpha\right) ( − cos α , − sin α )

The value of 2 ∫ sin ⁡ x d x sin ⁡ ( x − π 4 ) = \sqrt{2}\int_{ }^{ }\frac{\sin xdx}{\sin\left(x-\frac{\pi}{4}\right)}= 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ∫ sin ( x − 4 π ) sin x d x =

x + ln ⁡ ∣ sin ⁡ ( x − π 4 ) ∣ + c x+\ln\left|\sin\left(x-\frac{\pi}{4}\right)\right|+c x + ln ∣ ∣ ∣ sin ( x − 4 π ) ∣ ∣ ∣ + c

∫ d x cos ⁡ x − sin ⁡ x = \int_{ }^{ }\frac{dx}{\cos x-\sin x}= ∫ cos x − sin x d x =

1 2 ln ⁡ ∣ tan ⁡ ( x 2 + 3 π 8 ) ∣ + c \frac{1}{\sqrt{2}}\ln\left|\tan\left(\frac{x}{2}+\frac{3\pi}{8}\right)\right|+c 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> 1 ln ∣ ∣ ∣ ∣ tan ( 2 x + 8 3 π ) ∣ ∣ ∣ ∣ + c

2 ln ⁡ ∣ tan ⁡ ( x 2 + 3 π 8 ) ∣ + c \sqrt{2}\ln\left|\tan\left(\frac{x}{2}+\frac{3\pi}{8}\right)\right|+c 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ln ∣ ∣ ∣ ∣ tan ( 2 x + 8 3 π ) ∣ ∣ ∣ ∣ + c

2 ln ⁡ ∣ tan ⁡ ( x 2 − 3 π 8 ) ∣ + c \sqrt{2}\ln\left|\tan\left(\frac{x}{2}-\frac{3\pi}{8}\right)\right|+c 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ln ∣ ∣ ∣ ∣ tan ( 2 x − 8 3 π ) ∣ ∣ ∣ ∣ + c

∫ d x x ( 2 − x ) = \int_{ }^{ }\frac{dx}{\sqrt{x\ \left(2-\sqrt{x}\right)}}= ∫ x ( 2 − x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ) <path d="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167, 158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067 c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84, 175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71 c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3, -252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26 s76,-59,76,-59s76,-60,76,-60z M1001 80H40000v40H1012z"> d x =

− 2 ln ⁡ ∣ 2 − x ∣ + c -2\ln\left|2-\sqrt{x}\right|+c − 2 ln ∣ ∣ 2 − x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ∣ ∣ + c

2 ln ⁡ ∣ 2 + x ∣ + c 2\ln\left|2+\sqrt{x}\right|+c 2 ln ∣ ∣ 2 + x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ∣ ∣ + c

− 2 ln ⁡ ∣ 2 + x ∣ + c -2\ln\left|2+\sqrt{x}\right|+c − 2 ln ∣ ∣ 2 + x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ∣ ∣ + c

∫ sin ⁡ x sin ⁡ x − cos ⁡ x d x = \int_{ }^{ }\frac{\sin x}{\sin x-\cos x}dx\ = ∫ sin x − cos x sin x d x =

1 2 [ x + ln ⁡ ∣ sin ⁡ x − cos ⁡ x ∣ ] + c \frac{1}{2}\left[x+\ln\left|\sin x-\cos x\right|\right]+c 2 1 [ x + ln ∣ sin x − cos x ∣ ] + c

1 3 [ x + ln ⁡ ∣ sin ⁡ x − c o x ∣ ] + c \frac{1}{3}\left[x+\ln\left|\sin x-cox\right|\right]+c 3 1 [ x + ln ∣ sin x − c o x ∣ ] + c

1 3 [ x + ln ⁡ ∣ sin ⁡ x + c o x ∣ ] + c \frac{1}{3}\left[x+\ln\left|\sin x+cox\right|\right]+c 3 1 [ x + ln ∣ sin x + c o x ∣ ] + c

∫ ( a + x a − x + a − x a + x ) d x = \int_{ }^{ }\left(\frac{\sqrt{a+x}}{a-x}+\frac{\sqrt{a-x}}{a+x}\right)dx= ∫ ( a − x a + x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> + a + x a − x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ) d x =

2 a sin ⁡ − 1 ( x a ) + c 2a\sin^{-1}\left(\frac{x}{a}\right)+c 2 a sin − 1 ( a x ) + c

2 sin ⁡ − 1 ( x a ) + c 2\sin^{-1}\left(\frac{x}{a}\right)+c 2 sin − 1 ( a x ) + c

2 cos ⁡ − 1 ( x a ) + c 2\cos^{-1}\left(\frac{x}{a}\right)+c 2 cos − 1 ( a x ) + c

2 a cos ⁡ − 1 ( x a ) + c 2a\cos^{-1}\left(\frac{x}{a}\right)+c 2 a cos − 1 ( a x ) + c

∫ ( 1 + x − 1 x ) e x + 1 x d x = \int_{ }^{ }\left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx= ∫ ( 1 + x − x 1 ) e x + x 1 d x =

x e x + 1 x + c xe^{x+\frac{1}{x}}+c x e x + x 1 + c

( x + 1 ) e x + 1 x + c \left(x+1\right)e^{x+\frac{1}{x}}+c ( x + 1 ) e x + x 1 + c

− x e x + 1 x + c -xe^{x+\frac{1}{x}}+c − x e x + x 1 + c

− x e x + 1 x -xe^{x+\frac{1}{x}} − x e x + x 1

2 n c 2 + 2 n c 4 + . . . . + 2 n c 2 n = 2n_{c2}+2n_{c4}+....+2n_{c2n}= 2 n c 2 + 2 n c 4 + . . . . + 2 n c 2 n =

2 2 n − 1 − 1 2^{2n-1}-1 2 2 n − 1 − 1

2 2 n − 1 2^{2n}-1 2 2 n − 1

2 2 n − 1 2^{2n-1} 2 2 n − 1

If n is odd, n c 1 + n c 3 + n c 5 + . . . . . . . . . . n_{c1}+n_{c3}+n_{c5}+.......... n c 1 + n c 3 + n c 5 + . . . . . . . . . . . . . . + n c 2 [ n 2 ] − 1 = ....+nc_{2\left[\frac{n}{2}\right]-1}= . . . . + n c 2 [ 2 n ] − 1 = where [ ] denotes g.i.f

2 n − 1 − 1 2^{n-1}-1 2 n − 1 − 1

C 0 + 2. C 1 + 3. C 2 + . . . . . C_0+2.C_1+3.C_2+..... C 0 + 2 . C 1 + 3 . C 2 + . . . . . . . . + ( n + 1 ) . C n = ...+\left(n+1\right).C_n= . . . + ( n + 1 ) . C n =

( n + 2 ) . 2 n − 1 \left(n+2\right).2^{n-1} ( n + 2 ) . 2 n − 1

( n + 2 ) 2 n \left(n+2\right)2^n ( n + 2 ) 2 n

n . 2 n − 1 n.2^{n-1} n . 2 n − 1

c 1 c 0 + 2. c r c 1 + 3. c 3 c 2 + . . . . \frac{c_1}{c_0}+2.\ \frac{c_r}{c_1}+3.\ \frac{c_3}{c_2}+.... c 0 c 1 + 2 . c 1 c r + 3 . c 2 c 3 + . . . . . . . + n c n c n − 1 = ...+n\ \frac{c_n}{c_{n-1}}= . . . + n c n − 1 c n =

n ( n + 1 ) 2 \frac{n\left(n+1\right)}{2} 2 n ( n + 1 )

( n + 1 ) ( n + 2 ) 2 \frac{\left(n+1\right)\left(n+2\right)}{2} 2 ( n + 1 ) ( n + 2 )

n ( n − 1 ) 2 \frac{n\left(n-1\right)}{2} 2 n ( n − 1 )

n ( n + 2 ) 2 \frac{n\left(n+2\right)}{2} 2 n ( n + 2 )

n P r = n_{P_r}= n P r =

r ! n C r r!\ n_{C_r} r ! n C r

n C r − n C r − 1 n_{C_r}-n_{C_{r-1}} n C r − n C r − 1

n P r − 1 n_{P_{r-1}} n P r − 1

For a parallel combination of two springs of force constants k 1 k_1 k 1 and k 2 k_2 k 2 the equivalent force constant of the combination is

k 1 k 2 k 1 + k 2 \frac{k_1k_2}{k_1+k_2} k 1 + k 2 k 1 k 2

k 1 + k 2 k 1 k 2 \frac{k_1+k_2}{k_1k_2} k 1 k 2 k 1 + k 2

k 1 − k 2 k_1-k_2 k 1 − k 2

Work is said to be done by a force

if the force or component of the force is along the direction of the displacement

if the magnitude of the force is not equal to zero

A bob of mass m is suspended from the ceiling of a train moving with an acceleration ‘a’ as shown in the figure. Find the angle θ in equilibrium position.

tan ⁡ − 1 ( a g ) \tan^{-1}\left(\frac{a}{g}\right) tan − 1 ( g a )

tan ⁡ − 1 ( 2 a g ) \tan^{-1}\left(\frac{2a}{g}\right) tan − 1 ( g 2 a )

tan ⁡ − 1 ( a 2 g ) \tan^{-1}\left(\frac{a}{2g}\right) tan − 1 ( 2 g a )

tan ⁡ − 1 ( a 3 g ) \tan^{-1}\left(\frac{a}{3g}\right) tan − 1 ( 3 g a )

All surfaces are smooth in the following figure. Find F such that block remains stationary with respect to wedge.

( M + m ) g tan ⁡ θ \left(M+m\right)g\tan\theta ( M + m ) g tan θ

( M + m ) g sin ⁡ θ \left(M+m\right)g\sin\theta ( M + m ) g sin θ

2 ( M − m ) g tan ⁡ θ 2\left(M-m\right)g\tan\theta 2 ( M − m ) g tan θ

( M + m ) g cot ⁡ θ \left(M+m\right)g\cot\theta ( M + m ) g cot θ

A cage in which a bird is sitting is hung from a spring balance. After some time the bird starts flying in the cage. The reading of weight in the spring balance will

depends upon whether the bird flies up or down

The number of normal forces acting on a body depends on

number of points or surfaces of contact

The acceleration of the system is

a = ( m 2 − m 1 ( m 1 + m 2 ) ) g a=\left(\frac{m_2-m_1}{\left(m_1+m_2\right)}\right)g a = ( ( m 1 + m 2 ) m 2 − m 1 ) g

a = m 2 g m 1 + m 2 a=\frac{m_2g}{m_1+m_2} a = m 1 + m 2 m 2 g

a = ( m 1 g ( m 1 + m 2 ) ) a=\left(\frac{m_1g}{\left(m_1+m_2\right)}\right) a = ( ( m 1 + m 2 ) m 1 g )

a = ( 2 m 1 m 2 ( m 1 + m 2 ) ) g a=\left(\frac{2m_1m_2}{\left(m_1+m_2\right)}\right)g a = ( ( m 1 + m 2 ) 2 m 1 m 2 ) g

Statement-I: The work done by a friction on a body sliding down on inclined plane is positive. Statement-II: Work done is greater than zero if angle between force and displacement is acute.

Statement I is false, Statement II is true

Both Statement I and II are true

Both Statement I and II are false

Statement I is true, Statement II is false

Statement-I: Both a stretched spring and a compressed spring have potential energy Statement-II: Work is to be done against restoring force in each case of stretching and compressing of the spring.

Both Statement I and II are true

Both Statement I and II are false

Statement I is true, Statement II is false

Statement I is false, Statement II is true

The types of bonds present in $$CuSO_4,\ 5H_2O\ $$ are

Electrovalent, covalent, co-ordinate covalent and hydrogen bond

Electrovalent and co-ordinate covalent

Covalent and co-ordinate covalent

Statement 1: An atom with lone pair of electron may act as donor Statement 2 : $$BF_3,\ BCl_3$$ acts as donor species

Statement I is correct and statement II is incorrect

Both statement I and II are correct

Both statement I and II are incorrect

Statement I is incorrect and statement II is correct

Statement I : The ratio of lone pair of electron to bond pair of electron in valence shell of a molecule of $$IF_7$$ is 3:1 Statement II : In $$IF_7$$ I shows $$sp^3d^3$$ hybridisation to attain pentagonal bi-pyramid geometry

Both statement I and II are correct.

Both statement I and II are incorrect

Statement I is correct and statement II is incorrect.

Statement I is incorrect and statement II is correct.

10th IIT Cumulative - 5

Authored by Santosha Santosha Sir

Physics, Chemistry, Mathematics

10th Grade - Professional Development

Used 1+ times

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$\int_{ }^{ }\frac{dx}{\cos x+\sqrt{3}\sin x}=$

$\frac{1}{2}\ln\left|\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)\right|+c$

$\frac{1}{2}\ln\left|\tan\left(\frac{x}{2}-\frac{\pi}{12}\right)\right|+c$

$\frac{1}{2}\left[\ln\left|\tan\left(x-\frac{\pi}{12}\right)\right|\right]+c$

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10th IIT Cumulative - 5

∫x3x+1dx=\int_{ }^{ }\frac{x^3}{x+1}dx=∫​x+1x3​dx=

∫x32x+1dx=\int_{ }^{ }\frac{x^3}{2x+1}dx=∫​2x+1x3​dx=

∫(exln⁡a−ealn⁡x+3ealn⁡a)dx=\int_{ }^{ }\left(e^{x\ln a}-e^{a\ln x}+3e^{a\ln a}\right)dx=∫​(exlna−ealnx+3ealna)dx=

∫11−4x2dx=\int_{ }^{ }\frac{1}{\sqrt{1-4x^2}}dx=∫​1−4x2​1​dx=

∫sin⁡2x.cos⁡2xdx=\int_{ }^{ }\sin^2x.\cos^2xdx=∫​sin2x.cos2xdx=

∫(sin⁡6x+cos⁡6xtan⁡2x)sec⁡4xdx=\int_{ }^{ }\left(\frac{\sin^6x+\cos^6x}{\tan^2x}\right)\sec^4xdx=∫​(tan2xsin6x+cos6x​)sec4xdx=

If ∫sin⁡xsin⁡(x−α)dx =\int_{ }^{ }\frac{\sin x}{\sin\left(x-\alpha\right)}dx\ =∫​sin(x−α)sinx​dx = Ax+Blog⁡sin⁡(x−α)+C Ax+B\log\sin\left(x-\alpha\right)+C\ Ax+Blogsin(x−α)+C then (A,B)=then\ \left(A,B\right)=then (A,B)=

The value of 2∫sin⁡xdxsin⁡(x−π4)=\sqrt{2}\int_{ }^{ }\frac{\sin xdx}{\sin\left(x-\frac{\pi}{4}\right)}=2​∫​sin(x−4π​)sinxdx​=

∫dxcos⁡x−sin⁡x=\int_{ }^{ }\frac{dx}{\cos x-\sin x}=∫​cosx−sinxdx​=

∫dxcos⁡x+3sin⁡x=\int_{ }^{ }\frac{dx}{\cos x+\sqrt{3}\sin x}=∫​cosx+3​sinxdx​=

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$\int_{ }^{ }\frac{x^3}{x+1}dx=$

$\int_{ }^{ }\frac{x^3}{2x+1}dx=$

$\int_{ }^{ }\left(e^{x\ln a}-e^{a\ln x}+3e^{a\ln a}\right)dx=$

$\int_{ }^{ }\frac{1}{\sqrt{1-4x^2}}dx=$

$\int_{ }^{ }\sin^2x.\cos^2xdx=$

$\int_{ }^{ }\left(\frac{\sin^6x+\cos^6x}{\tan^2x}\right)\sec^4xdx=$

If $\int_{ }^{ }\frac{\sin x}{\sin\left(x-\alpha\right)}dx\ =$ $Ax+B\log\sin\left(x-\alpha\right)+C\$ $then\ \left(A,B\right)=$

The value of $\sqrt{2}\int_{ }^{ }\frac{\sin xdx}{\sin\left(x-\frac{\pi}{4}\right)}=$

$\int_{ }^{ }\frac{dx}{\cos x-\sin x}=$

$\int_{ }^{ }\frac{dx}{\cos x+\sqrt{3}\sin x}=$