# Current Affairs JEE Main & Advanced

#### Formulae for Sum, Difference of Inverse Trigonometric Function

(1) ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$;      If $x>0,y>0$ and $xy<1$     (2) ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$;     If $x>0,\,y>0$ and $xy>1$     (3) ${{\tan }^{-1}}x+{{\tan }^{-1}}y=-\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$;                 If $x<0,\,y<0$ and $xy>1$     (4) ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right)$;                             If $xy>-1$     (5) ${{\tan }^{-1}}x-{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right)$ ;                                    If $x>0,\,y<0$ and $xy<-1$     (6) ${{\tan }^{-1}}x-{{\tan }^{-1}}y=-\pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right)$;                   If $x<0,\,y>0$ and $xy<-1$     (7) ${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z={{\tan }^{-1}}\left[ \frac{x+y+z-xyz}{1-xy-yz-zx} \right]$     (8) ${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+..........+{{\tan }^{-1}}{{x}_{n}}$ $={{\tan }^{-1}}\left[ \frac{{{S}_{1}}-{{S}_{3}}+{{S}_{5}}-...........}{1-{{S}_{2}}+{{S}_{4}}-{{S}_{6}}+........} \right]$     where ${{S}_{k}}$ denotes the sum of the products of ${{x}_{1}},\,{{x}_{2}},........,{{x}_{n}}$ taken k  at a time.     (9) ${{\cot }^{-1}}x+{{\cot }^{-1}}y={{\cot }^{-1}}\frac{xy-1}{y+x}$     (10) ${{\cot }^{-1}}x-{{\cot }^{-1}}y={{\cot }^{-1}}\frac{xy+1}{y-x}$     (11) ${{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\{x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}\}$;     If $-1\le x,\,y\le 1$and${{x}^{2}}+{{y}^{2}}\le 1$ or if $xy<0$ and ${{x}^{2}}+{{y}^{2}}>1$     (12) ${{\sin }^{-1}}x+{{\sin }^{-1}}y=\pi -{{\sin }^{-1}}\{x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}\},$      If $0<x$,$y\le 1$ and ${{x}^{2}}+{{y}^{2}}>1$     (13) ${{\sin }^{-1}}x+{{\sin }^{-1}}y=-\pi -{{\sin }^{-1}}\{x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}\},$     If $-1\le x;\,y<0$ and ${{x}^{2}}+{{y}^{2}}>1$     (14) ${{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\{x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}}\},$     If $-1\le x;\,y\le 1$and${{x}^{2}}+{{y}^{2}}\le 1$if or $xy>0$ and${{x}^{2}}+{{y}^{2}}>1$.     (15) ${{\sin }^{-1}}x-{{\sin }^{-1}}y=\pi -{{\sin }^{-1}}\{x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}}\},$     If $0<x\le 1,\,-1\le y<0$ and ${{x}^{2}}+{{y}^{2}}>1$.     (16) ${{\sin }^{-1}}x-{{\sin }^{-1}}y=-\pi -{{\sin }^{-1}}\{x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}}\},$     If $-1\le x<0,\,0<y\le 1$ and ${{x}^{2}}+{{y}^{2}}>1$.     (17) ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\{xy-\sqrt{1-{{x}^{2}}}.\sqrt{1-{{y}^{2}}}\}$,     If $-1\le x,\,y\le 1$ and $x+y\ge 0$.     (18) ${{\cos }^{-1}}x+{{\cos }^{-1}}y=2\pi -{{\cos }^{-1}}\{xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}\}$,     If $-1\le x,\,y\le 1$ and $x+y\le 0$     (19) ${{\cos }^{-1}}x-{{\cos }^{-1}}y={{\cos }^{-1}}\{xy+\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}\},$      If $-1\le x,y\le 1,$  and $x\le y$.     (20) ${{\cos }^{-1}}x-{{\cos }^{-1}}y=-{{\cos }^{-1}}\{xy+\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}\},$     If $-1\le y\le 0,$ $0<x\le 1$ and $x\ge y$.

#### Properties of inverse trigonometric functions

(1) ${{\sin }^{-1}}(\sin \theta )=\theta$,   Provided that $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$,      ${{\cos }^{-1}}(\cos \theta )=\theta$,   Provided that $0\le \theta \le \pi$   ${{\tan }^{-1}}(\tan \theta )=\theta$,   Provided that $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$,               ${{\cot }^{-1}}(\cot \theta )=\theta$,   Provided that $0<\theta <\pi$   ${{\sec }^{-1}}(\sec \theta )=\theta$,  Provided that $0\le \theta <\frac{\pi }{2}$ or $\frac{\pi }{2}<\theta \le \pi$            $\text{cose}{{\text{c}}^{-1}}(\text{cosec}\theta \text{)}=\theta \text{,}$Provided that $-\frac{\pi }{2}\le \theta <0$or $0<\theta \le \frac{\pi }{2}$   (2) $\sin ({{\sin }^{-1}}x)=x,$ Provided that $-1\le x\le 1$,          $\cos ({{\cos }^{-1}}x)=x,$ Provided that $-1\le x\le 1$   tan $({{\tan }^{-1}}x)=x,$ Provided that $-\infty <x<\infty$   $\cot ({{\cot }^{-1}}x)=x,$ Provided that $-\infty <x<\infty$   $\sec ({{\sec }^{-1}}x)=x,$ Provided that $-\infty <x\le -1$ or $1\le x<\infty$   $\text{cosec }(\text{cose}{{\text{c}}^{\text{--1}}}x)=x,$Provided that $-\infty <x\le -1$ or $1\le x<\infty$   (3) ${{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x$,     ${{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}x$   ${{\tan }^{-1}}(-x)=-{{\tan }^{-1}}x$ ,    ${{\cot }^{-1}}(-x)=\pi -{{\cot }^{-1}}x$         ${{\sec }^{-1}}(-x)=\pi -{{\sec }^{-1}}x$,   $\text{cose}{{\text{c}}^{-1}}(-x)=-\text{cose}{{\text{c}}^{\text{--1}}}x$   (4) ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}$,    for all $x\in [-1,\,1]$   ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2}$,    for all $x\in R$ ${{\sec }^{-1}}x+\text{cose}{{\text{c}}^{\text{-1}}}x=\frac{\pi }{2}$,    for all $x\in (-\infty ,\,-1]\cup [1,\,\infty )$   (5) Principal values for inverse circular functions
Principal values for x ³ 0 Principal values for x <  0
$0\le {{\sin }^{-1}}x\le \frac{\pi }{2}$ $-\frac{\pi }{2}\le {{\sin }^{-1}}x<0$
$0\le {{\cos }^{-1}}x\le \frac{\pi }{2}$ $\frac{\pi }{2}<{{\cos }^{-1}}x\le \pi$
$0\le {{\tan }^{-1}}x<\frac{\pi }{2}$ $-\frac{\pi }{2}<{{\tan }^{-1}}x<0$
$0<{{\cot }^{-1}}x\le \frac{\pi }{2}$ $\frac{\pi }{2}<{{\cot }^{-1}}x<\pi$
$0\le {{\sec }^{-1}}x<\frac{\pi }{2}$ $\frac{\pi }{2}<{{\sec }^{-1}}x\le \pi$
$0<\text{cose}{{\text{c}}^{-1}}x\le \frac{\pi }{2}$ more...

#### Integration of Hyperbolic Functions

(1) $\int{\sinh x\,dx}=\cos \,\text{h}x\,+c$             (2) $\int{\cos \text{h}x\,dx=\sinh x+c}$     (3) $\int{\sec \,{{\text{h}}^{2}}x\,dx=\tan \,\text{h}\,x+c}$                   (4) $\int{\text{cosec}{{\text{h}}^{2}}x\,dx=-\cot \text{h}x+c}$     (5) $\int{\sec \,\text{h}x\,\tan \,\text{h}x\,dx=-\sec \text{h}x+c}$                       (6) $\int{\text{cosec}\,\text{h}x\,\cot \,\text{h}\,dx=-\text{cosec}\,\text{h}x+c}$

#### Integration of Trigonometric Functions

(1) Integral of the form $\int{si{{n}^{m}}x\,co{{s}^{n}}x\,dx}$: (i) To evaluate the integrals of the form $I=\int{{{\sin }^{m}}x\,{{\cos }^{n}}x\,dx,}$ where $m$ and $n$ are rational numbers.     (a) Substitute $\sin x=t,$ if $n$ is odd;     (b) Substitute cos x = t, if m is odd;     (c) Substitute $\tan x=t,$ if $m+n$ is a negative even integer; and     (d) Substitute $\cot x=t,$ if $\frac{1}{2}(n-1)$ is an integer.     (e) If $m$ and $n$ are rational numbers and $\left( \frac{m+n-2}{2} \right)$ is a negative integer, then substitution $\cos x=t$ or $\tan x=t$ is found suitable.     (ii) Integrals of the form $\int{R(\sin x,\,\cos x)\,dx,}$ where $R$ is a rational function of $\sin x$ and $\cos x,$ are transformed into integrals of a rational function by the substitution $\tan \frac{x}{2}=t,$where $-\pi <x<\pi .$  This is the so called universal substitution. Sometimes it is more convenient to make the substitution $\cot \frac{x}{2}=t$ for $0<x<2\pi .$     The above substitution enables us to integrate any function of the form $R(\sin x,\,\cos x).$ However, in practice, it sometimes leads to extremely complex rational function. In some cases, the integral can be simplified by :     (a) Substituting $\sin x=t,$ if the integral is of the form $\int{R(\sin x)\cos x\,dx}$.     (b) Substituting $+N(c\cos x-d\sin x)+P.$ if the integral is of the form $\int{R(\cos x)\sin x\,dx}$.     (c) Substituting tan $x=t$, i.e., $dx=\frac{dt}{1+{{t}^{2}}},$ if the integral is dependent only on $\tan x.$     (d) Substituting $\cos x=t$, if $R(-\sin x,\,\cos x)=-R(\sin x,\,\cos x)$     (e) Substituting $\sin x=t$, if $R(\sin x,-\,\cos x)=-R(\sin x,\,\cos x)$     (f) Substituting $\tan x=t$, if $R(-\sin x,-\,\cos x)=-R(\sin x,\,\cos x)$     (2) Reduction formulae for special cases     (i) $\int{{{\sin }^{n}}x\,dx=\frac{-\cos x\,.\,{{\sin }^{n-1}}x}{n}}+\frac{n-1}{n}\int{{{\sin }^{n-2}}x\,dx}$            (ii) $\int{{{\cos }^{n}}x\,dx=\frac{\sin x{{\cos }^{n-1}}x}{n}}+\frac{n-1}{n}\int{{{\cos }^{n-2}}x\,dx}$     (iii) $\int{{{\tan }^{n}}x\,dx=\frac{{{\tan }^{n-1}}x}{n-1}-\int{{{\tan }^{n-2}}}x\,dx}$     (iv) $\int{{{\cot }^{n}}x\,dx=\frac{-1}{n-1}{{\cot }^{n-1}}x-\int{{{\cot }^{n-2}}x\,dx}}$     (v) $\int{{{\sec }^{n}}x\,dx=\frac{1}{(n-1)}\left[ {{\sec }^{n-2}}x.\,\tan x+(n-2)\int{{{\sec }^{n-2}}}x\,dx \right]}$     (vi) $\int{\text{cose}{{\text{c}}^{n}}}xdx=\frac{1}{(n-1)}[-\text{cose}{{\text{c}}^{n-2}}x\cot x+(n-2)\int{\text{cose}{{\text{c}}^{n-2}}xdx}]$     (vii) $\int{{{\sin }^{p}}x{{\cos }^{q}}x\,dx=-\frac{{{\sin }^{q+1}}x.\,{{\cos }^{p-1}}x}{p+q}}$$\sin x=\frac{2\tan (x/2)}{1+{{\tan }^{2}}(x/2)}$     (viii) $\int{{{\sin }^{p}}x{{\cos }^{q}}x\,dx=\frac{{{\sin }^{p+1}}x\,.{{\cos }^{q-1}}x}{p+q}}$ $+\frac{p-1}{p+q}\int{{{\sin }^{p}}x.{{\cos }^{q-2}}x\,dx}$     (ix) $\int{\frac{dx}{{{({{x}^{2}}+k)}^{n}}}=\frac{x}{k(2n-2)\,{{({{x}^{2}}+k)}^{n-1}}}+\frac{(2n-3)}{k(2n-2)}}\,\int{\frac{dx}{{{({{x}^{2}}+k)}^{n-1}}}}$

#### Integration of Rational Functions by Using Partial Fractions

If the given function is in the form of fractions of two polynomials, then for its integration, decompose it into partial fractions (if possible). In the beginning chapters, we already discussed the decomposition of partial fractions.

#### Integrals of the form $\int{\frac{a sinx+b cosx}{c sinx+d cosx} }$and $\int{\frac{a\,sinx+bcos\,x+q}{c\,sinx+d\,cosx+r}}$

(1) Integrals of the form$\int{\frac{a sinx+b cos x}{c sinx+d\,\mathbf{cos} x}}\,dx$: Such rational functions of $\sin x$ and $\cos x$ may be integrated by expressing the numerator of the integrand as follows :     Numerator $=M$ (Diff. of denominator) $+N$ (Denominator)     i.e., $a\sin x+b\cos x=M\frac{d}{dx}(c\sin x+d\cos x)+N(c\sin x+d\cos x)$     The arbitary constants $M$ and $N$ are determined by comparing the coefficients of $\sin x$ and $\cos x$ from two sides of the above identity. Then, the given integral is     $I=\int{\frac{a\sin x+b\cos x}{c\sin x+d\cos x}}\,dx$     $=\int{\frac{M(c\cos x-d\sin x)+N(c\sin x+d\cos x)}{c\sin x+d\cos x}}\,dx$     $=M\int{\frac{c\cos x-d\sin x}{c\sin x+d\cos x}}\,dx+N\int{1dx}$     $=M\log |c\sin x+d\cos x|+Nx+c.$     (2) Integrals of the form $\int{\frac{a\,sinx+bcosx+q}{csinx+dcosx+r}}\,dx$ : To evaluate this type of integrals, we express the numerator as follows: Numerator $=M\text{(Denominator)}+N\text{(Differentiation}\,\text{of}\,\text{denominator)}+P$     i.e.,$(c\sin x+b\cos x+q)=M(c\sin x+d\cos x+r)$ $+N(c\cos x-d\sin x)+P.$     where M, N, P  are constants to be determined by comparing the coefficients of $\sin x,\,\cos x$ and constant term on both sides.     $\therefore \,\,\int{\frac{a\sin x+b\cos x+q}{c\sin x+d\cos x+r}}\,dx$     $=\int{M\,dx}+N\int{\frac{\text{Diff}\text{.}\,\text{of}\,\text{denominator}}{\text{Denominator}}\,dx}$$+\int{\frac{dx}{c\sin x-d\cos x+r}}$     $=Mx+N\log |\text{Denominator}|\,$ $+P\int{\frac{dx}{c\sin x+d\cos x+r}}$.

#### Integrals of the form $\int_{{}}^{{}}{\frac{dx}{a+b\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}x},\,\int_{{}}^{{}}{\frac{dx}{a+b\,\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}x},}}$$\int_{{}}^{{}}{\frac{dx}{a\,\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}x+b\,\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}x}\mathbf{,}}$$\int_{{}}^{{}}{\frac{dx}{{{\mathbf{(}a\,\sin x+b\,\mathbf{cos}\,x\mathbf{)}}^{\mathbf{2}}}}\mathbf{,}}\int_{{}}^{{}}{\frac{dx}{a\,+b\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}x+c\,\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}x}}$

To evaluate the above forms of integrals proceed as follows:     (1) Divide both the numerator and denominator by ${{\cos }^{2}}x.$     (2) Replace ${{\sec }^{2}}x$in the denominator, if any by $(1+{{\tan }^{2}}x).$     (3) Put $\tan x=t\,\,\,\Rightarrow \,\,{{\sec }^{2}}xdx=dt.$     (4) Now, evaluate the integral thus obtained, by the method discussed earlier.

#### Integrals of the Form $\int{\frac{dx}{a+bcosx+csinx}}\,,\,\,\int{\frac{dx}{asinx+bcosx}}$

(1) Integral of the form $\int{\frac{dx}{a+b\,cos\,x+csinx}}$: To evaluate such integrals, we put $b=r\cos \alpha$ and $c=r\sin \alpha .$     So that, ${{r}^{2}}={{b}^{2}}+{{c}^{2}}$ and $\alpha ={{\tan }^{-1}}\frac{c}{b}.$      $\therefore \,\,\,I=\int{\frac{dx}{a+r(\cos \alpha \,\cos x+\sin \alpha \sin x)}}$$=\int{\frac{dx}{a+r\cos (x-\alpha )}}$     Again, put $x-\alpha =t\,$$\Rightarrow \,\,dx=dt,$ we have $I=\int{\frac{dt}{a+r\cos t}}$     Which can be evaluated by the method discussed earlier.     (2) Integral of the form $\int{\frac{dx}{a\,sin\,x+b\,cos\,x}}$ : To evaluate this type of integrals we substitute $a=r\cos \theta ,$ $b=r\sin \theta$ and so $r=\sqrt{{{a}^{2}}+{{b}^{2}}},\,\,\alpha ={{\tan }^{-1}}\frac{b}{a}$ .     So, $\int{\frac{dx}{a\sin x+b\cos x}}=\frac{1}{r}\int{\frac{dx}{\sin (x+\alpha )}}=\frac{1}{r}\int{\text{cosec}(x+\alpha )dx}$     $=\frac{1}{r}\log \left| \tan \left( \frac{x}{2}+\frac{\alpha }{2} \right)\, \right|$$=\frac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\log \left| \tan \left( \frac{x}{2}+\frac{1}{2}{{\tan }^{-1}}\frac{b}{a} \right)\, \right|+c$     The integral of the above form can be evaluated by using $\cos x=\frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2}$ and $\sin x=\frac{2\tan (x/2)}{1+{{\tan }^{2}}(x/2)}$.

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